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1. • CommentRowNumber1.
   • CommentAuthorRichard Williamson
   • CommentTimeSep 29th 2018
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   Author: Richard Williamson
   Format: MarkdownItexHello all, I was wondering if one or two of you would be able to find the time to see if you think that the arguments on the following page, on my personal web, are correct? There are so many sharp minds amongst the readership here, that I'm sure if there is an error, one of you will make short work of finding it! [[richardwilliamson:Observations on prime divisors of odd integers in a range]] With these kinds of elementary arguments, everything has to be exactly right; the entire argument is likely to collapse if the smallest detail is wrong. Thus, though I have checked the argument several times myself, including after having slept on it, it is eminently possible that I have overlooked something, so do read with a critical eye :-). I am also thinking of making a little page, for completeness, on Bertrand's postulate, on the main nLab, which I can link to in my argument. Would this be OK (I ask because I don't know any category theoretic point of view on it at all)? Any feedback will be greatly appreciated!

   Hello all, I was wondering if one or two of you would be able to find the time to see if you think that the arguments on the following page, on my personal web, are correct? There are so many sharp minds amongst the readership here, that I’m sure if there is an error, one of you will make short work of finding it!

   Observations on prime divisors of odd integers in a range (richardwilliamson)

   With these kinds of elementary arguments, everything has to be exactly right; the entire argument is likely to collapse if the smallest detail is wrong. Thus, though I have checked the argument several times myself, including after having slept on it, it is eminently possible that I have overlooked something, so do read with a critical eye :-).

   I am also thinking of making a little page, for completeness, on Bertrand’s postulate, on the main nLab, which I can link to in my argument. Would this be OK (I ask because I don’t know any category theoretic point of view on it at all)?

   Any feedback will be greatly appreciated!

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 29th 2018
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   Author: Todd_Trimble
   Format: MarkdownItexI think it's absolutely fine to write on Bertrand's postulate.

   I think it’s absolutely fine to write on Bertrand’s postulate.

3. • CommentRowNumber3.
   • CommentAuthorRichard Williamson
   • CommentTimeSep 29th 2018
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   Author: Richard Williamson
   Format: MarkdownItexGreat! I'll do that now, I have a little time.

   Great! I’ll do that now, I have a little time.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 29th 2018
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   Author: Todd_Trimble
   Format: MarkdownItexI can try to go through this slowly, line by line. The argument seems a bit intricate. First, could I get you to clarify the following point? > Let $n \geq 6$ be an even integer, and let $k \geq 1$ be an integer. Let $S$ be a set of odd integers with at least $k$ elements such that $max(S) - min(S) \leq p_{k} - 3$, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. Then at least one of the following holds. > 1. There is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$. > 2. There is an odd integer $3 \leq t \lt min(S)$ and a prime $p$ such that $p \mid t$ and $p \gt p_{k}$. [snip] > Proof. We proceed by induction on $n$. When $n$ is equal to $6$, $8$, or $10$, 1. holds in each case for all possible $k$ and $S$. Let me just look at $n = 6$ for a moment. So here the second set of inequalities force $S = \{3\}$. We do seem to have the first inequality on $\max(S) - \min(S)$ for any prime $p_k$. But then to say that for any such $p_k$ there is $p \geq p_k$ such that $p \mid 3$ holds seems wrong (take $p_k = 5$). Did you mean to say $p \leq p_k$ instead?

   I can try to go through this slowly, line by line. The argument seems a bit intricate. First, could I get you to clarify the following point?

   Let $n \geq 6$ be an even integer, and let $k \geq 1$ be an integer. Let $S$ be a set of odd integers with at least $k$ elements such that $max(S) - min(S) \leq p_{k} - 3$, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. Then at least one of the following holds.

   1. There is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$.

   2. There is an odd integer $3 \leq t \lt min(S)$ and a prime $p$ such that $p \mid t$ and $p \gt p_{k}$. [snip]

   Proof. We proceed by induction on $n$. When $n$ is equal to $6$, $8$, or $10$, 1. holds in each case for all possible $k$ and $S$.

   Let me just look at $n = 6$ for a moment. So here the second set of inequalities force $S = \{3\}$. We do seem to have the first inequality on $\max(S) - \min(S)$ for any prime $p_k$. But then to say that for any such $p_k$ there is $p \geq p_k$ such that $p \mid 3$ holds seems wrong (take $p_k = 5$). Did you mean to say $p \leq p_k$ instead?

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 29th 2018
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   Author: Todd_Trimble
   Format: MarkdownItexOh, never mind. $S$ needs to have at least $k$ elements, which takes care of my question. Let me have at it again in a while.

   Oh, never mind. $S$ needs to have at least $k$ elements, which takes care of my question. Let me have at it again in a while.

6. • CommentRowNumber6.
   • CommentAuthorRichard Williamson
   • CommentTimeSep 29th 2018
   • (edited Sep 29th 2018)
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   Author: Richard Williamson
   Format: MarkdownItexThanks enormously, Todd, that would be absolutely perfect! Yes, the assumption that $S$ must have at least $k$ elements is crucial. There may be a little room to tweak the conditions around what currently is stated as $max(S) - min(S) \leq p_k - 3$, but let's see how it goes with that condition in its present form.

   Thanks enormously, Todd, that would be absolutely perfect!

   Yes, the assumption that $S$ must have at least $k$ elements is crucial. There may be a little room to tweak the conditions around what currently is stated as $max(S) - min(S) \leq p_k - 3$, but let’s see how it goes with that condition in its present form.

7. • CommentRowNumber7.
   • CommentAuthorRichard Williamson
   • CommentTimeSep 29th 2018
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   Author: Richard Williamson
   Format: MarkdownItexHi Todd, I have spotted a gap. I'll let you know if I can fix it; don't spend any time on the argument meanwhile! But thanks greatly for your willingness to work through it, it is uplifting!

   Hi Todd, I have spotted a gap. I’ll let you know if I can fix it; don’t spend any time on the argument meanwhile! But thanks greatly for your willingness to work through it, it is uplifting!

8. • CommentRowNumber8.
   • CommentAuthorRichard Williamson
   • CommentTimeSep 30th 2018
   • (edited Sep 30th 2018)
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexHello again, Todd, I've had another crack at [[richardwilliamson:Observations on prime divisors of odd integers in a range]]. If you would still be willing to go through it, I would be delighted if you could take a look. Theorem 2.1 may look outwardly quite different to its previous incarnation, but this is a bit deceptive; the fundamental ideas are the same, but it is very delicate to find a formulation of the theorem which allows the induction to go through. Again, I've proof read it myself without finding an error yet, but it is very easy to overlook something. The gap in the previous version was quite subtle and only in one tiny detail, though obvious in hindsight. And though the error in the proof was not easy to spot when considered purely abstractly, it was in fact very easy to construct a counterexample to the previous incarnation of Theorem 2.1. That should be much harder to do this time, because the current formulation is only a slight relaxation of something which I have checked to hold on the computer for all even integers up to about $10^6$.

   Hello again, Todd, I’ve had another crack at Observations on prime divisors of odd integers in a range (richardwilliamson). If you would still be willing to go through it, I would be delighted if you could take a look.

   Theorem 2.1 may look outwardly quite different to its previous incarnation, but this is a bit deceptive; the fundamental ideas are the same, but it is very delicate to find a formulation of the theorem which allows the induction to go through.

   Again, I’ve proof read it myself without finding an error yet, but it is very easy to overlook something. The gap in the previous version was quite subtle and only in one tiny detail, though obvious in hindsight. And though the error in the proof was not easy to spot when considered purely abstractly, it was in fact very easy to construct a counterexample to the previous incarnation of Theorem 2.1. That should be much harder to do this time, because the current formulation is only a slight relaxation of something which I have checked to hold on the computer for all even integers up to about $10^6$.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 30th 2018
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   Author: Todd_Trimble
   Format: MarkdownItexFirst, there must have been a subtle error in the link you provided, because it landed me on a page not found. Let's see if [this](https://ncatlab.org/richardwilliamson/show/Observations%20on%20prime%20divisors%20of%20odd%20integers%20in%20a%20range) works better. I'm sure you would agree that Theorem 2.1 has a lot of moving parts. There is no way I can digest the hypotheses as a gestalt; the only procedure seems to be to start with the proof and refer back to the hypotheses as needed. Let me start just with the first sentence, and see if I follow. In future posts I'll try battling my way through future parts of the proof. > Let $n \geq 16$ be an even integer, and let $k \geq 1$ be an integer. Suppose that for all integers $1 \leq i \leq k-1$, there is no prime $p$ such that $p \mid n - p_{i}$ and $p \geq p_{k}$. Let $S$ be a set of odd integers such that, for all integers $1 \leq i \leq k$, the set $S_{i} = \{ s \in S \mid s \geq n - (p_{i} + 6)\}$ has at least $i$ elements, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. If $k \gt 2$ and $n - 9 \in S$, suppose in addition that there does not exist a prime $p$ such that $p \mid n - 9$ and $p \geq p_{k}$. Then there is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$. > Proof. We proceed by induction on $k$. When $k=1$, it is clear that the theorem holds for any $S$ satisfying the conditions. Well, the hypothesis beginning with "Suppose that" is vacuously satisfied in the case $k=1$. For the next hypothesis, we are to suppose $S$ is a set of odd integers such that $3 \leq s \leq n-3$ for all $s \in S$, and such that $S_1 = \{s \in S: s \geq n-8\}$ has at least $1$ element. I can ignore the hypothesis "If $k \gt 2$ ..." We must conclude that there exists $s \in S$ such that $p \mid S$ and $p \geq 2$. Yep, seems undeniable since there is $s \in S$ such that $s \geq 8$, and this has some prime divisor $p$. So the proof easily passes this stage of the inquisition. :-) I'll look at the next part later.

   First, there must have been a subtle error in the link you provided, because it landed me on a page not found. Let’s see if this works better.

   I’m sure you would agree that Theorem 2.1 has a lot of moving parts. There is no way I can digest the hypotheses as a gestalt; the only procedure seems to be to start with the proof and refer back to the hypotheses as needed. Let me start just with the first sentence, and see if I follow. In future posts I’ll try battling my way through future parts of the proof.

   Let $n \geq 16$ be an even integer, and let $k \geq 1$ be an integer. Suppose that for all integers $1 \leq i \leq k-1$, there is no prime $p$ such that $p \mid n - p_{i}$ and $p \geq p_{k}$. Let $S$ be a set of odd integers such that, for all integers $1 \leq i \leq k$, the set $S_{i} = \{ s \in S \mid s \geq n - (p_{i} + 6)\}$ has at least $i$ elements, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. If $k \gt 2$ and $n - 9 \in S$, suppose in addition that there does not exist a prime $p$ such that $p \mid n - 9$ and $p \geq p_{k}$. Then there is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$.

   Proof. We proceed by induction on $k$. When $k=1$, it is clear that the theorem holds for any $S$ satisfying the conditions.

   Well, the hypothesis beginning with “Suppose that” is vacuously satisfied in the case $k=1$.

   For the next hypothesis, we are to suppose $S$ is a set of odd integers such that $3 \leq s \leq n-3$ for all $s \in S$, and such that $S_1 = \{s \in S: s \geq n-8\}$ has at least $1$ element.

   I can ignore the hypothesis “If $k \gt 2$ …”

   We must conclude that there exists $s \in S$ such that $p \mid S$ and $p \geq 2$. Yep, seems undeniable since there is $s \in S$ such that $s \geq 8$, and this has some prime divisor $p$. So the proof easily passes this stage of the inquisition. :-)

   I’ll look at the next part later.

10. • CommentRowNumber10.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 30th 2018
    • (edited Sep 30th 2018)
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    Author: Richard Williamson
    Format: MarkdownItexThis is absolutely perfect Todd, thanks very much! > First, there must have been a subtle error in the link you provided, because it landed me on a page not found. My apologies! Thanks for pointing it out, fixed now. > I’m sure you would agree that Theorem 2.1 has a lot of moving parts. Absolutely. I could provide some discussion of it, but actually, if you are able to continue in this perfect way in checking the argument, it may even be better for the moment not to have any intuition for it, and just to take it at its cold face value! > and such that $S_1 = \{s \in S: s \geq n-8\}$ has at least $1$ element. There was a slight typo in your comment here: it should be $s \geq n - 9$, not $s \geq n - 8$. It doesn't matter at this point, of course, but one has to be very careful with these inequalities later. > So the proof easily passes this stage of the inquisition. :-) Hehe, great :-)!

    This is absolutely perfect Todd, thanks very much!

    First, there must have been a subtle error in the link you provided, because it landed me on a page not found.

    My apologies! Thanks for pointing it out, fixed now.

    I’m sure you would agree that Theorem 2.1 has a lot of moving parts.

    Absolutely. I could provide some discussion of it, but actually, if you are able to continue in this perfect way in checking the argument, it may even be better for the moment not to have any intuition for it, and just to take it at its cold face value!

    and such that $S_1 = \{s \in S: s \geq n-8\}$ has at least $1$ element.

    There was a slight typo in your comment here: it should be $s \geq n - 9$, not $s \geq n - 8$. It doesn’t matter at this point, of course, but one has to be very careful with these inequalities later.

    So the proof easily passes this stage of the inquisition. :-)

    Hehe, great :-)!

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 30th 2018
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    Author: Todd_Trimble
    Format: MarkdownItexLet me just ask about that typo. I was starting the enumeration of primes with $p_1 = 2$. You're not?

    Let me just ask about that typo. I was starting the enumeration of primes with $p_1 = 2$. You’re not?

12. • CommentRowNumber12.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 30th 2018
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    Author: Richard Williamson
    Format: MarkdownItexI should probably state this explicitly in the statement of the theorem, but I'm working only with odd primes, i.e. $p_{1} = 3$. This is stated in the last sentence of the introduction, but as I say, I should make it even clearer.

    I should probably state this explicitly in the statement of the theorem, but I’m working only with odd primes, i.e. $p_{1} = 3$. This is stated in the last sentence of the introduction, but as I say, I should make it even clearer.

13. • CommentRowNumber13.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 30th 2018
    • (edited Sep 30th 2018)
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    Author: Richard Williamson
    Format: MarkdownItexThus, to be fully careful, where you wrote > Yep, seems undeniable since there is $s \in S$ such that $s \geq 8$, and this has some prime divisor $p$ we should actually observe more precisely that since $s$ is odd, it has some _odd_ prime divisor.

    Thus, to be fully careful, where you wrote

    Yep, seems undeniable since there is $s \in S$ such that $s \geq 8$, and this has some prime divisor $p$

    we should actually observe more precisely that since $s$ is odd, it has some odd prime divisor.

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 30th 2018
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    Author: Todd_Trimble
    Format: MarkdownItexOkay, that's what I figured. Thanks. More later.

    Okay, that’s what I figured. Thanks. More later.

15. • CommentRowNumber15.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 30th 2018
    • (edited Sep 30th 2018)
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    Author: Richard Williamson
    Format: MarkdownItexFantastic! Off to bed now, but I'll reply to any comments/questions as soon as I get the chance.

    Fantastic! Off to bed now, but I’ll reply to any comments/questions as soon as I get the chance.

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 1st 2018
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    Author: Todd_Trimble
    Format: MarkdownItex> Let $n \geq 16$ be an even integer, and let $k \geq 1$ be an integer. Suppose that for all integers $1 \leq i \leq k-1$, there is no prime $p$ such that $p \mid n - p_{i}$ and $p \geq p_{k}$. Let $S$ be a set of odd integers such that, for all integers $1 \leq i \leq k$, the set $S_{i} = \{ s \in S \mid s \geq n - (p_{i} + 6)\}$ has at least $i$ elements, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. If $k \gt 2$ and $n - 9 \in S$, suppose in addition that there does not exist a prime $p$ such that $p \mid n - 9$ and $p \geq p_{k}$. Then there is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$. > Proof. We proceed by induction on $k$. When $k=1$, it is clear that the theorem holds for any $S$ satisfying the conditions. > Suppose that $k=2$. If the theorem does not hold for some $S$ satisfying the conditions, the only possibility is that $S_{2}$ has at least two elements $s$ and $s'$, both of which are powers of $3$. Suppose first that neither $s$ nor $s'$ is prime. Then the difference between $s$ and $s'$ is at least $3^{2} \cdot (3 - 1) = 18$. But the difference between $s$ and $s'$ is in fact at most $(n-3) - \left(n - (5 + 6) \right) = 8$, and thus we have a contradiction. > Suppose instead that at least one of $s$ or $s'$ is prime, that is to say, equal to $3$. We then must have that $3 = min(S_{2}) \geq n - (5 + 6) = n - 11$. But since $n \geq 16$, we have that $n - 11 \gt 3$. We thus, once more, have a contradiction. > We conclude that the theorem does hold when $k=2$. Okay, for the moment I'm ignoring your proof text (which I'll come back to in a moment) just to write down what I can extract from your hypotheses for the case $k = 2$. The first hypothesis asks us to suppose that there is no prime $p \geq 5$ such that $p \mid n-3$. The next hypothesis has two parts, in addition to the condition that $3 \leq s \leq n-3$ for all $s \in S$: * There is an element $s \in S$ with $s \geq n-9$, * There are two elements $s_1, s_2 \in S$ such that $s_1,s_2 \geq n-11$. Evidently, in this case the two conditions coalesce into the second one. Okay, let's carry on. We are trying to prove that some prime $p \geq 5$ divides some $s \in S$. "If the theorem does not hold for some $S$ satisfying the conditions, the only possibility is that $S_{2}$ has at least two elements $s$ and $s'$, both of which are powers of $3$." Yes, that seems to be the case, remembering of course that $S$ consists only of odd integers. "Suppose first that neither $s$ nor $s'$ is prime. Then the difference between $s$ and $s'$ is at least $3^2 \cdot (3 - 1) =18$." Yes, because they are both powers of $3$ that are greater than $3$. "But the difference between $s$ and $s'$ is in fact at most $(n-3) - \left(n - (5 + 6) \right) = 8$, and thus we have a contradiction." Yes. "Suppose instead that at least one of $s$ or $s'$ is prime, that is to say, equal to $3$. We then must have that $3 = \min(S_{2}) \geq n - (5 + 6) = n - 11$. But since $n \geq 16$, we have that $n - 11 \gt 3$. We thus, once more, have a contradiction." Yes, I agree. "We conclude that the theorem does hold when $k=2$." Yes, looks good to me. More later.

    Let $n \geq 16$ be an even integer, and let $k \geq 1$ be an integer. Suppose that for all integers $1 \leq i \leq k-1$, there is no prime $p$ such that $p \mid n - p_{i}$ and $p \geq p_{k}$. Let $S$ be a set of odd integers such that, for all integers $1 \leq i \leq k$, the set $S_{i} = \{ s \in S \mid s \geq n - (p_{i} + 6)\}$ has at least $i$ elements, and such that $max(S) \leq n - 3$ and $min(S) \geq 3$. If $k \gt 2$ and $n - 9 \in S$, suppose in addition that there does not exist a prime $p$ such that $p \mid n - 9$ and $p \geq p_{k}$. Then there is an $s \in S$ and a prime $p$ such that $p \mid s$ and $p \geq p_{k}$.

    Proof. We proceed by induction on $k$. When $k=1$, it is clear that the theorem holds for any $S$ satisfying the conditions.

    Suppose that $k=2$. If the theorem does not hold for some $S$ satisfying the conditions, the only possibility is that $S_{2}$ has at least two elements $s$ and $s'$, both of which are powers of $3$. Suppose first that neither $s$ nor $s'$ is prime. Then the difference between $s$ and $s'$ is at least $3^{2} \cdot (3 - 1) = 18$. But the difference between $s$ and $s'$ is in fact at most $(n-3) - \left(n - (5 + 6) \right) = 8$, and thus we have a contradiction.

    Suppose instead that at least one of $s$ or $s'$ is prime, that is to say, equal to $3$. We then must have that $3 = min(S_{2}) \geq n - (5 + 6) = n - 11$. But since $n \geq 16$, we have that $n - 11 \gt 3$. We thus, once more, have a contradiction.

    We conclude that the theorem does hold when $k=2$.

    Okay, for the moment I’m ignoring your proof text (which I’ll come back to in a moment) just to write down what I can extract from your hypotheses for the case $k = 2$. The first hypothesis asks us to suppose that there is no prime $p \geq 5$ such that $p \mid n-3$. The next hypothesis has two parts, in addition to the condition that $3 \leq s \leq n-3$ for all $s \in S$:

    • There is an element $s \in S$ with $s \geq n-9$,

    • There are two elements $s_1, s_2 \in S$ such that $s_1,s_2 \geq n-11$.

    Evidently, in this case the two conditions coalesce into the second one. Okay, let’s carry on. We are trying to prove that some prime $p \geq 5$ divides some $s \in S$.

    “If the theorem does not hold for some $S$ satisfying the conditions, the only possibility is that $S_{2}$ has at least two elements $s$ and $s'$, both of which are powers of $3$.” Yes, that seems to be the case, remembering of course that $S$ consists only of odd integers.

    “Suppose first that neither $s$ nor $s'$ is prime. Then the difference between $s$ and $s'$ is at least $3^2 \cdot (3 - 1) =18$.” Yes, because they are both powers of $3$ that are greater than $3$.

    “But the difference between $s$ and $s'$ is in fact at most $(n-3) - \left(n - (5 + 6) \right) = 8$, and thus we have a contradiction.” Yes.

    “Suppose instead that at least one of $s$ or $s'$ is prime, that is to say, equal to $3$. We then must have that $3 = \min(S_{2}) \geq n - (5 + 6) = n - 11$. But since $n \geq 16$, we have that $n - 11 \gt 3$. We thus, once more, have a contradiction.” Yes, I agree.

    “We conclude that the theorem does hold when $k=2$.” Yes, looks good to me.

    More later.

17. • CommentRowNumber17.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 1st 2018
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    Author: Richard Williamson
    Format: MarkdownItexGreat! Thanks so much Todd, this is really perfect.

    Great! Thanks so much Todd, this is really perfect.

18. • CommentRowNumber18.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 1st 2018
    • (edited Oct 1st 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexI think I just noticed a tiny, but not serious, gap, in the following line. > Suppose first that there is a prime $p''$ such that $p'' \mid n - 9$ and $p'' \geq p_{k-1}$. If $p'' \gt p_{k-1}$, this contradicts one of our assumptions As stated, I don't think it does, because the assumption about $n-9$ is made under the hypothesis that $n-9$ belongs to $S$ (i.e. the original $S$ in the statement of Theorem 2.1), which it may not do. For now, just assume that the assumption about $n-9$ is made independently of $S$. This slightly affects the generality of Corollary 2.2 and Theorem 2.3, but not to a very great degree. Moreover, I think one will be able to treat the cases where the assumption about $n-9$ does not hold if everything else goes through.

    I think I just noticed a tiny, but not serious, gap, in the following line.

    Suppose first that there is a prime $p''$ such that $p'' \mid n - 9$ and $p'' \geq p_{k-1}$. If $p'' \gt p_{k-1}$, this contradicts one of our assumptions

    As stated, I don’t think it does, because the assumption about $n-9$ is made under the hypothesis that $n-9$ belongs to $S$ (i.e. the original $S$ in the statement of Theorem 2.1), which it may not do. For now, just assume that the assumption about $n-9$ is made independently of $S$. This slightly affects the generality of Corollary 2.2 and Theorem 2.3, but not to a very great degree. Moreover, I think one will be able to treat the cases where the assumption about $n-9$ does not hold if everything else goes through.

19. • CommentRowNumber19.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 1st 2018
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    Author: Richard Williamson
    Format: MarkdownItexHmm, actually one cannot quite simply make that assumption independently of $S$ either without tweaking some other parts of the argument. I'll look into it later; please do not spend more time on it in the meantime, Todd, I'll let you know when it's in good shape again.

    Hmm, actually one cannot quite simply make that assumption independently of $S$ either without tweaking some other parts of the argument. I’ll look into it later; please do not spend more time on it in the meantime, Todd, I’ll let you know when it’s in good shape again.

20. • CommentRowNumber20.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 1st 2018
    • (edited Oct 1st 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexThe argument should be in a stable state again now, Todd, if you are able to continue when convenient. Apologies for the disruption. I believe I have addressed the issues of #18 and #19 now. The part of the argument which you have already verified is unchanged. However, the conclusion is slightly stronger (we require now that $s \neq n - 9$), so to make this part of the argument (i.e. when $k=1$ or $k=2$) still go through, I have strengthened the hypotheses when $k=1$ and $k=2$ slightly. Otherwise the hypotheses are the same as before, except that I have removed the one involving $k \gt 2$ and $n-9$, and have instead, as already mentioned, strengthened the conclusion. In fact, the argument for $k \gt 2$ is almost unchanged as well, actually simplified slightly. So if you had begun checking that part of the argument already before I notified about the gap in #18, your work should hopefully not have been wasted. Corollary 2.2 and Theorem 2.3 are entirely unchanged.

    The argument should be in a stable state again now, Todd, if you are able to continue when convenient. Apologies for the disruption. I believe I have addressed the issues of #18 and #19 now.

    The part of the argument which you have already verified is unchanged. However, the conclusion is slightly stronger (we require now that $s \neq n - 9$), so to make this part of the argument (i.e. when $k=1$ or $k=2$) still go through, I have strengthened the hypotheses when $k=1$ and $k=2$ slightly.

    Otherwise the hypotheses are the same as before, except that I have removed the one involving $k \gt 2$ and $n-9$, and have instead, as already mentioned, strengthened the conclusion.

    In fact, the argument for $k \gt 2$ is almost unchanged as well, actually simplified slightly. So if you had begun checking that part of the argument already before I notified about the gap in #18, your work should hopefully not have been wasted.

    Corollary 2.2 and Theorem 2.3 are entirely unchanged.

21. • CommentRowNumber21.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 2nd 2018
    • (edited Oct 2nd 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexI don't know if you've found time to take a look yet, Todd (if you're still willing; I understand if not!), but I just wished to mention by way of encouragement that I have been through the latest version of the argument, namely that discussed in #20, numerous times, and it definitely looks correct to me. Of course one can sometimes be blinded when looking at one's own work, so checking by yourself or anybody else will still be enormously appreciated, but I usually find an error in my work within a day or two of rumination if there is one!

    I don’t know if you’ve found time to take a look yet, Todd (if you’re still willing; I understand if not!), but I just wished to mention by way of encouragement that I have been through the latest version of the argument, namely that discussed in #20, numerous times, and it definitely looks correct to me. Of course one can sometimes be blinded when looking at one’s own work, so checking by yourself or anybody else will still be enormously appreciated, but I usually find an error in my work within a day or two of rumination if there is one!

22. • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 2nd 2018
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexI do plan on looking at it, Richard; Tuesdays are just very busy days for me. I'm glad you're encouraged! But you know and I know that you're making a bold claim. :-)

    I do plan on looking at it, Richard; Tuesdays are just very busy days for me. I’m glad you’re encouraged! But you know and I know that you’re making a bold claim. :-)

23. • CommentRowNumber23.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 2nd 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItex> I do plan on looking at it, Richard Fantastic! > Tuesdays are just very busy days for me. No problem at all, I am extremely grateful for the time you have already given, and you should absolutely feel no obligation to look any further at it; as you plan for the moment to do so, which is very generous of you, you should definitely not feel rushed! > But you know and I know that you’re making a bold claim. :-) Indeed :-). It would be remarkable if there were not a significant error somewhere. Immediately after having written #21, it was brought to my attention that there _was_ an oversight, in one tiny place I had indeed been completely blind to. Namely, when $k=3$, it was possible that the induction hypothesis for $S''$ might not be satisfied, because my requirement that there be at least two elements other than $n-9$ in $S_{2}$ might not hold. I had checked this extremely carefully for $S'$, but completely overlooked it for $S''$. For the moment, I have fixed this by removing the problematic hypothesis, and instead imposing some restrictions on $n$. This reduces the generality of Corollary 2.2 and Theorem 2.3 a bit, but they are still more than general enough to be interesting :-). The proof in the case $k=2$ is now different (very simple), but otherwise the argument is unchanged from before. I am hopeful that there might be a different fix which avoids reducing the generality, but I'll need to think more about that. It might be getting a bit difficult to keep track of which comments in this thread refer to which revisions. 1) The current version, with the latest fix, is [Revision 15](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/15). 2) The previous version, where the error with $k=3$ occurred, and to which #20 and #21 were referring, is [Revision 12](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/12). 3) The version before that, to which #8 to #19 were referring, is [Revision 8](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/8).

    I do plan on looking at it, Richard

    Fantastic!

    Tuesdays are just very busy days for me.

    No problem at all, I am extremely grateful for the time you have already given, and you should absolutely feel no obligation to look any further at it; as you plan for the moment to do so, which is very generous of you, you should definitely not feel rushed!

    But you know and I know that you’re making a bold claim. :-)

    Indeed :-). It would be remarkable if there were not a significant error somewhere.

    Immediately after having written #21, it was brought to my attention that there was an oversight, in one tiny place I had indeed been completely blind to. Namely, when $k=3$, it was possible that the induction hypothesis for $S''$ might not be satisfied, because my requirement that there be at least two elements other than $n-9$ in $S_{2}$ might not hold. I had checked this extremely carefully for $S'$, but completely overlooked it for $S''$.

    For the moment, I have fixed this by removing the problematic hypothesis, and instead imposing some restrictions on $n$. This reduces the generality of Corollary 2.2 and Theorem 2.3 a bit, but they are still more than general enough to be interesting :-). The proof in the case $k=2$ is now different (very simple), but otherwise the argument is unchanged from before.

    I am hopeful that there might be a different fix which avoids reducing the generality, but I’ll need to think more about that.

    It might be getting a bit difficult to keep track of which comments in this thread refer to which revisions.

    1) The current version, with the latest fix, is Revision 15.

    2) The previous version, where the error with $k=3$ occurred, and to which #20 and #21 were referring, is Revision 12.

    3) The version before that, to which #8 to #19 were referring, is Revision 8.

24. • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeOct 3rd 2018
    • (edited Oct 3rd 2018)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> But you know and I know that you’re making a bold claim. :-) Maybe for the rest of us, one of you could briefly comment on what's at stake here. What is bold about the claim? (I suppose we are speaking about theorem 2.3 at the bottom [here](https://ncatlab.org/richardwilliamson/show/Observations+on+prime+divisors+of+odd+integers+in+a+range).) I gather the concluding sentence of the theorem claims that primes are not far away, but I have no feeling for how strong the assumption clause of the statement is. It all looks very, what's the word, peculiarly particular. How do you even come up with the idea of such a statement?

    But you know and I know that you’re making a bold claim. :-)

    Maybe for the rest of us, one of you could briefly comment on what’s at stake here. What is bold about the claim?

    (I suppose we are speaking about theorem 2.3 at the bottom here.)

    I gather the concluding sentence of the theorem claims that primes are not far away, but I have no feeling for how strong the assumption clause of the statement is. It all looks very, what’s the word, peculiarly particular. How do you even come up with the idea of such a statement?

25. • CommentRowNumber25.
    • CommentAuthorUrs
    • CommentTimeOct 3rd 2018
    • PermaLink
    Author: Urs
    Format: MarkdownItexOh, I see now. It is something close to the Goldbach cojecture. All right.

    Oh, I see now. It is something close to the Goldbach cojecture. All right.

26. • CommentRowNumber26.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 3rd 2018
    • (edited Oct 3rd 2018)
    • PermaLink
    Author: Richard Williamson
    Format: TextYes, exactly, so the odds are overwhelming that there is a significant error somewhere :-). Just now I've spotted a further small gap. Will explain later. Please do not spend time on it until then, Todd (or anybody else!).

    Yes, exactly, so the odds are overwhelming that there is a significant error somewhere :-).
    
    Just now I've spotted a further small gap. Will explain later. Please do not spend time on it until then, Todd (or anybody else!).
27. • CommentRowNumber27.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 3rd 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexHi Todd and anybody else interested, I addressed the gap mentioned in #26 earlier today. The latest version is at [[richardwilliamson:Observations on prime divisors of odd integers in a range]], and is [Revision 19](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/19). If you are able to take a look when you get a chance, that would be fantastic. An important aspect of the latest version is that the role of $n-9$, which previously seemed rather too magical, now is very clear. An insight suddenly came to me which felt like exactly the missing piece. The claims are now back in full generality, not in the restricted generality of #23. The argument is still very close to that which you began checking in #9 - #16, Todd. However, the statement of Theorem 2.1 has been tweaked a bit; since #9 - #16, a couple of hypotheses have been removed, and the conclusion has been significantly tweaked. Thus, although the arguments when $k=1$ and $k=2$ are very similar to how they were before, it might be worth casting your eye over them again. [Edit: the gap in [Revision 15](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/15), by the way, that I noticed in #26, was that I had not handled the case that the assumption that there is no integer $1 \leq i \leq k-1$ such that $n - p_{i}$ is divisible by a prime $p \geq p_{k-1}$ does not hold. For indeed it may not hold: even though we know that there is no $p \geq p_{k}$ with this property, it is possible that $n - p_{i}$ could be equal to $p_{k-1}$ for some $i$. And my argument did not treat this case.]

    Hi Todd and anybody else interested, I addressed the gap mentioned in #26 earlier today. The latest version is at Observations on prime divisors of odd integers in a range (richardwilliamson), and is Revision 19.

    If you are able to take a look when you get a chance, that would be fantastic.

    An important aspect of the latest version is that the role of $n-9$, which previously seemed rather too magical, now is very clear. An insight suddenly came to me which felt like exactly the missing piece.

    The claims are now back in full generality, not in the restricted generality of #23.

    The argument is still very close to that which you began checking in #9 - #16, Todd. However, the statement of Theorem 2.1 has been tweaked a bit; since #9 - #16, a couple of hypotheses have been removed, and the conclusion has been significantly tweaked. Thus, although the arguments when $k=1$ and $k=2$ are very similar to how they were before, it might be worth casting your eye over them again.

    [Edit: the gap in Revision 15, by the way, that I noticed in #26, was that I had not handled the case that the assumption that there is no integer $1 \leq i \leq k-1$ such that $n - p_{i}$ is divisible by a prime $p \geq p_{k-1}$ does not hold. For indeed it may not hold: even though we know that there is no $p \geq p_{k}$ with this property, it is possible that $n - p_{i}$ could be equal to $p_{k-1}$ for some $i$. And my argument did not treat this case.]

28. • CommentRowNumber28.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 3rd 2018
    • (edited Oct 3rd 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexRe #24 and #25: > I have no feeling for how strong the assumption clause of the statement is. As I think you spotted afterwards, in its latest form, Theorem 2.3 would be the (strong) Goldbach conjecture in full generality. I should note that the proofs of Corollary 2.2 and Theorem 2.3 have already been checked to be correct, except for the first paragraph of Corollary 2.2. So everything hangs on Theorem 2.1 (and the first paragraph of Corollary 2.2). > How do you even come up with the idea of such a statement? Much hard work :-). More seriously, I am happy to provide some discussion as to how one might arrive at something vaguely like Theorem 2.1, but I'll leave it for the moment, because I think it is probably best for now just to try the verify the proof as-is, in its austerity. Going from the vague idea of Theorem 2.1 to the present versions has taken many months of effort (though I have not worked on it continuously, I just occasionally come back to it). Over the course of working on it, I basically have acquired a sort of small collection of tools ('mini-proofs' that I can carry out); the question is to try to assemble them in such a way that an inductive argument can be carried out. I feel I'm getting closer, but I'm of course perfectly aware of how remarkable it would be if this holds up (and thus how unlikely it is)!

    Re #24 and #25:

    I have no feeling for how strong the assumption clause of the statement is.

    As I think you spotted afterwards, in its latest form, Theorem 2.3 would be the (strong) Goldbach conjecture in full generality. I should note that the proofs of Corollary 2.2 and Theorem 2.3 have already been checked to be correct, except for the first paragraph of Corollary 2.2. So everything hangs on Theorem 2.1 (and the first paragraph of Corollary 2.2).

    How do you even come up with the idea of such a statement?

    Much hard work :-). More seriously, I am happy to provide some discussion as to how one might arrive at something vaguely like Theorem 2.1, but I’ll leave it for the moment, because I think it is probably best for now just to try the verify the proof as-is, in its austerity. Going from the vague idea of Theorem 2.1 to the present versions has taken many months of effort (though I have not worked on it continuously, I just occasionally come back to it). Over the course of working on it, I basically have acquired a sort of small collection of tools (’mini-proofs’ that I can carry out); the question is to try to assemble them in such a way that an inductive argument can be carried out. I feel I’m getting closer, but I’m of course perfectly aware of how remarkable it would be if this holds up (and thus how unlikely it is)!

29. • CommentRowNumber29.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI see. Last time I looked the last theorem had an assumption about a few numbers below the given one being or not being divisible by $3^2$, 5, 7, 11. I was wondering what to make of such an assumption. But now I see it has disappeared.

    I see. Last time I looked the last theorem had an assumption about a few numbers below the given one being or not being divisible by $3^2$, 5, 7, 11. I was wondering what to make of such an assumption. But now I see it has disappeared.

30. • CommentRowNumber30.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 4th 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexYes, this assumption was the reduction in generality which was briefly introduced in the previous version, [Revision 15](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/15) (discussed in #23), and then removed again in the latest version. The assumption came out of the proof of the $k=2$ step, but with the tweaked conclusion of Theorem 2.1 in the latest version, it was no longer necessary. For the record, the assumption would have had the effect of roughly halving the generality of Theorem 2.3.

    Yes, this assumption was the reduction in generality which was briefly introduced in the previous version, Revision 15 (discussed in #23), and then removed again in the latest version. The assumption came out of the proof of the $k=2$ step, but with the tweaked conclusion of Theorem 2.1 in the latest version, it was no longer necessary. For the record, the assumption would have had the effect of roughly halving the generality of Theorem 2.3.

31. • CommentRowNumber31.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks. Just out of curiosity: What would be some striking consequences of a proof of the Goldbach conjecture? Any other major statements that depend on it? In the wake of the recent non-proof of the Riemann hypthesis I saw some discussion of possible effects any such proof might or might not have on encryption standards. If I understood well, then a proof of the Riemann hypothesis would actually not have any effect on such practical matters while a _dis_proof might be more interesting, for practical implications. I gather the point was that the truth of the Riemann hypothesis implies, or would imply, that current algorithms for finding large primes (which apparently proceed by picking any one large number and then searching for primes nearby) are guaranteed to succeed in some nicely bounded time interval. Anything similar to say about the Goldbach conjecture? (Just out of curiosity.)

    Thanks. Just out of curiosity: What would be some striking consequences of a proof of the Goldbach conjecture? Any other major statements that depend on it?

    In the wake of the recent non-proof of the Riemann hypthesis I saw some discussion of possible effects any such proof might or might not have on encryption standards. If I understood well, then a proof of the Riemann hypothesis would actually not have any effect on such practical matters while a _dis_proof might be more interesting, for practical implications. I gather the point was that the truth of the Riemann hypothesis implies, or would imply, that current algorithms for finding large primes (which apparently proceed by picking any one large number and then searching for primes nearby) are guaranteed to succeed in some nicely bounded time interval.

    Anything similar to say about the Goldbach conjecture? (Just out of curiosity.)

32. • CommentRowNumber32.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 4th 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexGood question! What you write about the Riemann hypothesis makes sense. I actually do not know of any consequences of the Goldbach conjecture except for the fact that it implies [[Bertrand's postulate]], and I am using Bertrand's postulate in my argument, so even that would not be a consequence of my argument, if the argument holds up :-)! I am just interested in it for its own sake. I do think, though, that the Goldbach conjecture is very interesting in the respect that my feeling is that one _has_ to use elementary methods if one is to prove it; I don't think sophisticated machinery is likely to help. This obviously goes completely against the grain of modern mathematics. So one consequence of a proof of the kind I am trying to give might be that it might make people reflect a bit on this kind of thing. It is also expected that a proof of the Goldbach conjecture may lead to progress on the twin primes conjecture, but I have not thought about that; and I equally do not know of any significant consequences of the twin primes conjecture!

    Good question! What you write about the Riemann hypothesis makes sense.

    I actually do not know of any consequences of the Goldbach conjecture except for the fact that it implies Bertrand’s postulate, and I am using Bertrand’s postulate in my argument, so even that would not be a consequence of my argument, if the argument holds up :-)! I am just interested in it for its own sake.

    I do think, though, that the Goldbach conjecture is very interesting in the respect that my feeling is that one has to use elementary methods if one is to prove it; I don’t think sophisticated machinery is likely to help. This obviously goes completely against the grain of modern mathematics. So one consequence of a proof of the kind I am trying to give might be that it might make people reflect a bit on this kind of thing.

    It is also expected that a proof of the Goldbach conjecture may lead to progress on the twin primes conjecture, but I have not thought about that; and I equally do not know of any significant consequences of the twin primes conjecture!

33. • CommentRowNumber33.
    • CommentAuthorUrs
    • CommentTimeOct 4th 2018
    • PermaLink
    Author: Urs
    Format: MarkdownItex> I don't think sophisticated machinery is likely to help. This obviously goes completely against the grain of modern mathematics. So one consequence of a proof of the kind I am trying to give might be that it might make people reflect a bit on this kind of thing. For what it's worth, I don't believe one has to or should take sides in this dichotomy between the general abstract and the particular. One wants both. The broad overarching theory in order to see where on the map we are, and then the nitty-gritty lemmas to catalogize the flora and fauna at that spot. But let's see, there must be something about the Goldbach conjecture as a statement about [[Spec(Z)]]. In that language it says that every function on $Spec(\mathbb{Z})$ that vanishes at the closed point $(2)$ is the sum of two of the _coordinate functions_ $p_1, p_2 \in \mathcal{O}( Spec(\mathbb{Z}) )$ that vanish at $(p_1)$, $(p_2)$. Hm, need to think about that. But, sorry, let me not distract you (and Todd) from scrutinizing the proof, first. :-)

    I don’t think sophisticated machinery is likely to help. This obviously goes completely against the grain of modern mathematics. So one consequence of a proof of the kind I am trying to give might be that it might make people reflect a bit on this kind of thing.

    For what it’s worth, I don’t believe one has to or should take sides in this dichotomy between the general abstract and the particular. One wants both. The broad overarching theory in order to see where on the map we are, and then the nitty-gritty lemmas to catalogize the flora and fauna at that spot.

    But let’s see, there must be something about the Goldbach conjecture as a statement about Spec(Z). In that language it says that every function on $Spec(\mathbb{Z})$ that vanishes at the closed point $(2)$ is the sum of two of the coordinate functions $p_1, p_2 \in \mathcal{O}( Spec(\mathbb{Z}) )$ that vanish at $(p_1)$, $(p_2)$. Hm, need to think about that.

    But, sorry, let me not distract you (and Todd) from scrutinizing the proof, first. :-)

34. • CommentRowNumber34.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 4th 2018
    • (edited Oct 4th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItex> For what it’s worth, I don’t believe one has to or should take sides in this dichotomy between the general abstract and the particular. One wants both. Absolutely, I 100% agree :-). I wasn't thinking so much here about the level of abstraction. For instance, whilst I really know little about it, sieve theory certainly strikes me as sophisticated in the sense I had in mind, though I don't know a nice conceptual framework for it at a reasonable level of abstraction. I was thinking more about the sociological aspect that many mathematicians are a bit disdainful of elementary attempts to prove things; it might be nice to have an example which shows that this can still be fruitful. > But, sorry, let me not distract you (and Todd) from scrutinizing the proof, first. :-) No problem! I have occasionally speculated a tiny bit about how one might try to categorify the Goldbach conjecture, but have not come up with anything interesting so far.

    For what it’s worth, I don’t believe one has to or should take sides in this dichotomy between the general abstract and the particular. One wants both.

    Absolutely, I 100% agree :-). I wasn’t thinking so much here about the level of abstraction. For instance, whilst I really know little about it, sieve theory certainly strikes me as sophisticated in the sense I had in mind, though I don’t know a nice conceptual framework for it at a reasonable level of abstraction. I was thinking more about the sociological aspect that many mathematicians are a bit disdainful of elementary attempts to prove things; it might be nice to have an example which shows that this can still be fruitful.

    But, sorry, let me not distract you (and Todd) from scrutinizing the proof, first. :-)

    No problem!

    I have occasionally speculated a tiny bit about how one might try to categorify the Goldbach conjecture, but have not come up with anything interesting so far.

35. • CommentRowNumber35.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 4th 2018
    • PermaLink
    Author: Richard Williamson
    Format: TextAs ever, I have spotted a tiny but non-trivial gap in the latest version as well. Will explain later when I have more time. Thus don't proceed with the proof-checking for the moment, Todd (or anybody else doing so).

    As ever, I have spotted a tiny but non-trivial gap in the latest version as well. Will explain later when I have more time. Thus don't proceed with the proof-checking for the moment, Todd (or anybody else doing so).
36. • CommentRowNumber36.
    • CommentAuthoruser 170039
    • CommentTimeOct 6th 2018
    • (edited Oct 6th 2018)
    • PermaLink
    Author: user 170039
    Format: MarkdownItexDoes anyone here know how in the following version of the proof the following line holds, "...We have that $S_{k-1}$ satisfies the conditions of the theorem for $k-1$. Suppose first that $n-9$ does not belong to $S_{k-1}$. By induction, either there is an integer $1\le i\le k-1$ such that $n-p_i$ does not belong to $S_{k-1}$, in which case $n-p_i$ does not belong to $S$ and the theorem holds..."?

    Does anyone here know how in the following version of the proof the following line holds, “…We have that $S_{k-1}$ satisfies the conditions of the theorem for $k-1$. Suppose first that $n-9$ does not belong to $S_{k-1}$. By induction, either there is an integer $1\le i\le k-1$ such that $n-p_i$ does not belong to $S_{k-1}$, in which case $n-p_i$ does not belong to $S$ and the theorem holds…”?

37. • CommentRowNumber37.
    • CommentAuthorRichard Williamson
    • CommentTimeOct 7th 2018
    • (edited Oct 7th 2018)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexHi, thanks very much for your interest and for your question! If I understand your question correctly, say for the proof attempt in [Revision 25](https://ncatlab.org/richardwilliamson/revision/Observations+on+prime+divisors+of+odd+integers+in+a+range/25), we have that $n - p_{i} \geq n - (p_{k-1} + 6)$, and that $S_{k-1}$ consists of all elements of $S$ which are greater than or equal to $n - (p_{k-1} + 6)$. So if $n - p_{i}$ does not belong to $S_{k-1}$, then since the inequality defining $S_{k-1}$ and which $n-p_{i}$ satisfies is the same, the only possibility is that $n - p_{i}$ does not belong to $S$. The error in the proof comes later, in the following line. > By induction, we deduce that there is an $s'' \in S'' \setminus \{ n- 9 \}$ and a prime $p''$ such that $p'' \mid s''$ and $p'' \geq p_{k-1}$. Induction gives us only that _either_ this holds _or_ there is an $n - p_{i}$ does not belong to $S''$, and of course $s'$ could be equal to $n - p_{i}$ for some $i$, in which case induction does not give us anything useful. In retrospect, it is sort of clear to me that the proof cannot work: I think one has to somehow make use of an $s$ in $S_{k}$ but not in $S_{k-1}$. What I have done now, in the latest version at [[richardwilliamson:Observations on prime divisors of odd integers in a range]] is to give a proof which is definitely correct, but which is dependent on a hypothesis $H$. As explained in Remark 2.5, I can find a $H$ which satisfies three of the four required properties, but not the fourth. I have tried numerous variations on $H$, and variations on the logic (e.g. in all the attempts mentioned in this thread), but so far have not found anything that quite works, though one can come very close. If anyone has any insight into $H$, e.g. if you can give an argument that it is impossible to find such a $H$, or if you have a suggestion for a good choice to explore, I would be delighted to hear your thoughts! I myself remain optimistic that this approach can work.

    Hi, thanks very much for your interest and for your question!

    If I understand your question correctly, say for the proof attempt in Revision 25, we have that $n - p_{i} \geq n - (p_{k-1} + 6)$, and that $S_{k-1}$ consists of all elements of $S$ which are greater than or equal to $n - (p_{k-1} + 6)$. So if $n - p_{i}$ does not belong to $S_{k-1}$, then since the inequality defining $S_{k-1}$ and which $n-p_{i}$ satisfies is the same, the only possibility is that $n - p_{i}$ does not belong to $S$.

    The error in the proof comes later, in the following line.

    By induction, we deduce that there is an $s'' \in S'' \setminus \{ n- 9 \}$ and a prime $p''$ such that $p'' \mid s''$ and $p'' \geq p_{k-1}$.

    Induction gives us only that either this holds or there is an $n - p_{i}$ does not belong to $S''$, and of course $s'$ could be equal to $n - p_{i}$ for some $i$, in which case induction does not give us anything useful. In retrospect, it is sort of clear to me that the proof cannot work: I think one has to somehow make use of an $s$ in $S_{k}$ but not in $S_{k-1}$.

    What I have done now, in the latest version at Observations on prime divisors of odd integers in a range (richardwilliamson) is to give a proof which is definitely correct, but which is dependent on a hypothesis $H$. As explained in Remark 2.5, I can find a $H$ which satisfies three of the four required properties, but not the fourth. I have tried numerous variations on $H$, and variations on the logic (e.g. in all the attempts mentioned in this thread), but so far have not found anything that quite works, though one can come very close.

    If anyone has any insight into $H$, e.g. if you can give an argument that it is impossible to find such a $H$, or if you have a suggestion for a good choice to explore, I would be delighted to hear your thoughts! I myself remain optimistic that this approach can work.

38. • CommentRowNumber38.
    • CommentAuthorRichard Williamson
    • CommentTimeApr 10th 2020
    • (edited Apr 10th 2020)
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexJust a quick note that after a long time of not working on this, I had a bit of inspiration, and have written up an argument. I have already notified Todd over email. If anybody else would be willing to go through the proofs of Proposition 2.4 and Theorem 2.6 carefully, it would be enormously appreciated. To try to reduce possibility of an error, I have on this occasion gone carefully back through the proof, and written up a step-by-step justification of each assertion. This is Scholium 2.5. [[richardwilliamson:Observations on prime divisors of odd integers in a range]] I should perhaps say that the note which is referenced has been checked to be correct by Ben Green amongst others, so it should be possible to focus on the correctness of this actual page.

    Just a quick note that after a long time of not working on this, I had a bit of inspiration, and have written up an argument. I have already notified Todd over email. If anybody else would be willing to go through the proofs of Proposition 2.4 and Theorem 2.6 carefully, it would be enormously appreciated. To try to reduce possibility of an error, I have on this occasion gone carefully back through the proof, and written up a step-by-step justification of each assertion. This is Scholium 2.5.

    Observations on prime divisors of odd integers in a range (richardwilliamson)

    I should perhaps say that the note which is referenced has been checked to be correct by Ben Green amongst others, so it should be possible to focus on the correctness of this actual page.