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I want to show that there are infinitely many primes $p$ such that $p = 9 \pmod {10}$.

First, I can see that 19 is one of them. Assume there are finitely many, i.e., 19, $p_1, p_2 , \cdots , p_k$. Let $P = 19p_1 p_2 \cdots p_k$ and $N =4P^2 -5.$ I want to show that all prime divisors of $N$ are congruent modulo 1 or 4 mod 5 and $N=9 \pmod{10}$.

Thank you so much.

Any help will be appreciated.

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  • $\begingroup$ Step 1: $p=9\pmod{10}$ if and only if $p=9\pmod{2}$ and $p=9\pmod{5}$. This simplifies the problem, as all primes except 2 satisfy the first condition. $\endgroup$ – vadim123 Apr 19 '16 at 4:18
  • $\begingroup$ @vadim123: Please tell why all prime divisors of $N$ are congruent to 1 or 4 mod5 $\endgroup$ – Surojit Apr 19 '16 at 4:21
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    $\begingroup$ math.stackexchange.com/questions/373750/… $\endgroup$ – vadim123 Apr 19 '16 at 4:37
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It is enough to show that there are infinitely many primes of the form $5k-1$.

Let $n\ge 2$, and let $N=5(n!)^2-1$. We first show that every prime divisor of $N$ is of the form $5k\pm 1$.

Suppose that $p$ is a prime divisor of $N$. Then $5$ is a quadratic residue of $p$. A simple Legendre symbol calculation shows that this forces $p\equiv \pm 1\pmod{5}$.

Finally, the prime divisors of $N$ cannot be all of the form $5k+1$, else their product $N$ would be, but it isn't. So $N$ has a prime divisor of the form $5k-1$.

Any prime divisor of $5(n!)^2-1$ must be greater than $n$. So we have shown that for any $n$ there is a prime of the form $5k-1$ which is greater than $n$.

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