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I want to prove:

There are infinitely many primes $p$ with $p\equiv 1 \mod 4$.

My lecture notes say this can be proven by considering the numbers $(n!)^2+1$ and using that they satisfy $(n!)^2+1\equiv 1\mod 4$.

I have two questions:

  • How to conclude the claim from this fact?
  • Why is the fact true?
MyNameIs
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    $-1$ is a quadratic residue modulo any prime factor $p$ of $(n!)^2+1$. Therefore $p\equiv1\pmod4$. If there were only finitely many such primes what would happen if $n$ were the largest of that lot? – Jyrki Lahtonen Jul 19 '16 at 12:21
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    see this question – n1000 Jul 19 '16 at 12:49
  • Why is $p\equiv 1\mod 4$ if $p$ is a prime factor of $(n!)^2+1$, @JyrkiLahtonen ? – MyNameIs Jul 20 '16 at 10:05
  • Is the following argument OK? $p$ divides $(n!)^2+1$, hence $(n!)^2\equiv -1 \mod p$ and therefore $\left(\frac{-1}{p} \right)=1$. This implies, by Euler's lemma, that $p\equiv 1 \mod 4$. – MyNameIs Jul 20 '16 at 10:10
  • So we found a prime $p$ which is of the form $4k+1$. Now we repeat the procedure for $n'=(p!)^2+1$. This gives us a prime $p'$ of the form $4l+1$. This prime must be bigger than $p$ because otherwise it would divide $1$. – MyNameIs Jul 20 '16 at 10:12
  • That's the idea. Alternatively you can do a contrapositive argument. If there were only finitely many primes of this type, there would be a largest such prime... – Jyrki Lahtonen Jul 20 '16 at 10:25
  • Have you seen the proof of infinitely many primes? If so then using the hint the proof of this statement should follow similarly. – tigs May 20 '25 at 18:16

1 Answers1

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Below is a proof that there are infinitely many primes $mod(1,4)$ using Dirichlet method:

Define a mapping system $\chi_0$ such that

$$\boxed{\sum_p \frac{\chi_0(p)}{p} = \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \frac{1}{19} + ... = \infty}$$

since the sum of reciprocals of primes diverge.

Define another mapping system $\chi_1$ such that

$$\boxed{\sum_p \frac{\chi_1(p)}{p} = - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} - \frac{1}{19} + ... = O(1)}$$

by comparison with the regular series

$$- \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \frac{1}{17} - \frac{1}{19} + ... = \frac{\pi}{4} - 1$$

and using

$$\sum_p \frac{\chi_1(p)}{p} = \log \left(\frac{\pi}{4} - 1\right) + O(1) $$

Then

$$\sum_{p=mod(1,4)} \frac{1}{p} = \frac{1}{5} + \frac{1}{13} + \frac{1}{17} + ...$$

$$=\frac{1}{2}\left(\sum_p \frac{\chi_0(p)}{p} + \sum_p \frac{\chi_1(p)}{p}\right)$$ $$ = \frac{1}{2}\left(\infty + O(1)\right)$$

Hence

$$\boxed{\sum_{p=mod(1,4)} \frac{1}{p} = \infty}$$

and so there are infinitely many primes $mod (1,4)$.

Furthermore, for analysis use

$$\log \sum_{n} \frac{\chi(n)}{n}$$

$$=\log\prod_{p} \frac{1}{1 - \frac{\chi(p)}{p}}$$

by product of unique prime factorization

$$=-\sum_p \log \left(1 - \frac{\chi(p)}{p}\right)$$

$$=\sum_p \left(\frac{\chi(p)}{p} + \frac{\chi^2(p)}{2p^2} + \frac{\chi^3(p)}{3p^3} + ... \right)$$

by Maclaurin series expansion

$$=\sum_p \frac{\chi(p)}{p} + \sum_p \left(\frac{\chi^3(p)}{2p^2} + \frac{\chi^3(p)}{3p^3} + ... \right)$$

$$=\sum_p \frac{\chi(p)}{p} + O(1)$$

Hence

$$\boxed{\sum_p \frac{\chi(p)}{p} =\log \sum_{n} \frac{\chi(n)}{n} + O(1) }$$

where the right hand side is preferable for use in analysis than the left hand side.

E.g. proof that there are infinitely many primes mod (3, 16) using Dirichlet method:

Define 8 mapping systems such that

$$\begin{array} {|r|r|}\hline mod(n,16) & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\ \hline \chi_0(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \chi_1(n) & -1 & 1 & 1 & -1 & -1 & 1 & 1 & -1 \\ \hline \chi_2(n) & -i & 1 & 1 & -i & i & -1 & -1 & i \\ \hline \chi_3(n) & i & 1 & 1 & i & -i & -1 & -1 & -i \\ \hline \chi_4(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\ \hline \chi_5(n) & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 \\ \hline \chi_6(n) & -i & 1 & -1 & i & i & -1 & 1 & -i \\ \hline \chi_7(n) & i & 1 & -1 & -i & -i & -1 & 1 & i \\ \hline \end{array}$$

Then

$$\sum_\chi \left(\sum_p \frac{\chi(p)}{p} \right) $$ $$= \sum_p \frac{1}{p}\left(\sum_\chi \chi(p) \right)$$ $$= \sum_p \frac{1}{p} \left( \chi_0(p) + \chi_1(p) + \chi_2(p) + \chi_3(p) + \chi_4(p) + \chi_5(p) + \chi_6(p) + \chi_7(p)\right)$$

$$ = \sum_{p=mod(3,16)} \frac{1}{p} (8)$$

since contributions from all other primes cancel out completely (i.e. a sieve) by inspecting the table.

Hence

$$ \boxed{\sum_{p=mod(3,16)} \frac{1}{p} = \frac{1}{8}\sum_\chi \sum_p \frac{\chi(p)}{p}}$$

$$ = \frac{1}{8}\sum_\chi \left(\log \sum_{n} \frac{\chi(n)}{n} + O(1) \right)$$

$$= \frac{1}{8} \left( \log \sum_{n} \frac{\chi_0(n)}{n}+ \log \sum_{n} \frac{\chi_1(n)}{n}+ \log \sum_{n} \frac{\chi_2(n)}{n}+ \log \sum_{n} \frac{\chi_3(n)}{n}+ \log \sum_{n} \frac{\chi_4(n)}{n}+ \log \sum_{n} \frac{\chi_5(n)}{n}+ \log \sum_{n} \frac{\chi_6(n)}{n}+ \log \sum_{n} \frac{\chi_7(n)}{n} \right) + O(1)$$

$$= \frac{1}{8} \left( \log \sum_{n} \frac{\chi_0(n)}{n}+ O(1) + O(1)+ O(1)+ O(1)+ O(1)+ O(1)+ O(1)\right) + O(1)$$

$$= \frac{1}{8} \left(\log \sum_{n} \frac{1}{n} \right)+ O(1)$$

since $\chi_0(n)=1$

$$ = \infty$$

since $\sum_{n} \frac{1}{n} = \infty$.

Therefore

$$\boxed{\sum_{p=mod(3,16)} \frac{1}{p} = \infty}$$

and so there are infinitely many primes $mod (3,16)$.

James
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