Primes congruent to 1 modulo n

Question

I was wondering whether the following statement is true, and if so, how to prove it:

for all $n\in\mathbb{N}$, there exist prime numbers $p$ and $q$ such that $p\equiv1$ (mod $n$) and $q\equiv-1$ (mod $n$)

Feels like it should be true and that it should be fairly easy to prove, but I don't have any idea how. Could someone please shine some light on this for me?

Dirichlet's theorem on primes in arithmetic progressions tells us that there will in fact be infinitely many such primes. — Geoffrey Trang, Jan 21 '21 at 15:39
Thanks. Is there a simple proof for the existence of at least 1? — t-tough, Jan 21 '21 at 15:47
There a relatively simple way to prove there are infinitely many $p$ using the cyclotomic polynomials, roughly like Euclid’s proof of infinitely name primes. I don’t know of any way for proving even one $q$ exists. — Thomas Andrews, Jan 21 '21 at 15:58
This answer shows the case of $p$: https://math.stackexchange.com/a/1160778/7933 — Thomas Andrews, Jan 21 '21 at 19:13
Dirichlet proved that $an+b$ produces infinite many primes if and only if $a$ and $b$ are coprime, but this proof is not easy. This proof implies your statement. — Peter, Jan 22 '21 at 09:00

sirous · Answer 1 · 2021-01-24T07:23:52.943

Comment: You can construct such primes; tank an arbitrary number like n= 7 we can have:

$p=2\times7-1=13$ ⇒ $p=13\equiv -1 \mod (2)$

$q=6\times 7+1=43$ ⇒ $q=43 \equiv 1 \mod 7$

In fact we must have:

$p=k\cdot n+1$

$q=t\cdot n-1$

From second relation we have:

$n=\frac{q+1}t$

Plugging this in first relation we get:

$p=\frac k t (q+1)+1=\frac kt q+\frac kt +1$

Let $\frac kt=a$ then:

$p=a q + a+1$

This equation has infinite solutions for p, q for certain value of a, among them primes for p and q. The condition is $k>t ; t|k$.

Primes congruent to 1 modulo n

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