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.

  • $\begingroup$ 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, 2016 at 4:18
  • $\begingroup$ @vadim123: Please tell why all prime divisors of $N$ are congruent to 1 or 4 mod5 $\endgroup$
    – Surojit
    Apr 19, 2016 at 4:21
  • 1
    $\begingroup$ math.stackexchange.com/questions/373750/… $\endgroup$
    – vadim123
    Apr 19, 2016 at 4:37

1 Answer 1


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 $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.