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Suppose there are only finitely many primes of the form $25m+7$. Let $p_1,\ldots, p_r$ be such primes and let $M = (5p_1\ldots p_r)^2 + 7$. Then $N$ has form $25m + 7$, and is not divisible by any $p_i$. Let $p$ be any prime dividing $M$. Then $(5p_1\ldots p_r)^2 \equiv -7 \mod{p}$, but this means that the number $−7$ is a square$\mod p$,which is only possible if $p \equiv 7 \mod 25$. This contradicts the fact that no $p_i$ divides $M$. Therefore there must be infinitely of that form.

Does this prove my claim? I was following a layout of a proof for primes of the from $4n+1$, and tried to modify it.

Update: the claim that $-7$ is a square $\mod p$,whenever $p\equiv 7 \mod 25$ does not hold so some other claim is needed

mandella
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    "which is only possible if $p\equiv 7 \pmod{25}$" -- why is this true? – vadim123 Mar 14 '18 at 17:50
  • e.g. $2^2\equiv -7\pmod{11}$, but $11$ is not congruent to $7$, modulo $25$. – vadim123 Mar 14 '18 at 17:51
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    $-7$ is a square modulo $11$. – Wojowu Mar 14 '18 at 17:52
  • Actually, I mentioned that I was following a layout of another proof - so I am not sure about the answer – mandella Mar 14 '18 at 17:55
  • Of course this follows from Diriclet's Theorem. Is there any reason to think this is substantially easier than the general case? – lulu Mar 14 '18 at 18:11
  • @lulu I see it follows from that theorem but I would like to have a proof of it – mandella Mar 14 '18 at 19:05
  • Of course there may be a clever trick that works for this case, but at some point surely it is easier to just understand the proof of Dirichlet's Theorem. – lulu Mar 14 '18 at 21:17
  • @lulu But I want to find the trick .. maybe if i consider the product ofprimes or smth.. – mandella Mar 17 '18 at 10:09
  • Well, again, it isn't clear to me that there even is a sensible trick. I mean, something significantly easier than the proof of Dirichlet. If you want to press further on it, I'd start by trying to show that there are infinitely many primes of the form $5n+2$. Your desired claim is stronger than that, of course. – lulu Mar 17 '18 at 11:59
  • @Lulu I think i can show that, but I don't know how to relate it to my exercise – mandella Mar 17 '18 at 12:01
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    Again, it would be a major step towards what you want. If a given problem is too hard it is a very good idea to weaken the question slightly and see if you can prove that. Standard practice. To me, the $5n+2$ case already looks hard. Just saying "I want to find a trick" isn't a sensible way to proceed. – lulu Mar 17 '18 at 12:10
  • I think lulu is right; one should use Dirichlet's theorem, see here. – Dietrich Burde Mar 18 '18 at 15:23

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The logical structure of your proof is fine, but there is one huge claim that you have stated without proof:

the number $−7$ is a square$\mod p$, which is only possible if $p \equiv 7 \mod 25$.

This claim is in fact not true. For instance, $-7$ is a square mod $2$, or mod $11$, even though $2$ and $11$ are not $7$ mod $25$. More generally, by quadratic reciprocity, if $p\neq 2,7$ is a prime, then $-7$ is a square mod $p$ iff $p$ is a square mod $7$; that is, iff $p$ is $1,2,$ or $4$ mod $7$.

Eric Wofsey
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  • I understand. I followed another proof and I see it does not makes sense. Do you have any idea how to fix this? – mandella Mar 14 '18 at 19:59