Prove that series $(10n+4)^2+1$ contains infinitly many composite numbers

Question

Given sequence is $$4^2+1,\ 14^2+1,\ 24^2+1,\ 34^2+1...$$ How to show that there are infinite amount of composite numbers?

Answer 1

dI can provide a list of numbers that follow such a form, but are composite and infinite. However, they are not exhaustive.

$$(10n+4)^2 + 1 \equiv 0 \mod p$$ $$100n^2 + 80 n + 16 \equiv -1 \mod p$$ $$20n(5n+4) \equiv -17 \mod p$$

Here, we can simply assume that $p = 17$ to find a list of numbers of such a form that are divisible by $17$

$$20n(5n+4) \equiv 0 \mod 17$$

Therefore $$20n \equiv 0 \mod 17$$ Or $$5n \equiv13 \mod 17$$

Therefore, $n =0 + 17k$ or $n = 6 + 17k$

However, $n = 0$ gives $17$ ,which is prime, so we can rewrite this as

If $n = 17 + 17k$ or $6+17k$, $k \in \mathbb{Z}$, $k \ge 0$, then $f(n) = (10n+4)^2 + 1$ is divisible by $17$, which gives us an infinite amount of composite numbers

Answer 2

If you take any prime $p$ such that $p\mid (10n+4)^2+1$ then $p\mid (10(n+kp)+4)^2+1$ because $$(10(n+kp)+4)^2+1=(10n+4+kp)^2+1=(10n+4)^2+2kp(10n+4)+k^2p^2+1=(10n+4)^2+1+p(2k(10n+4)+k^2p)$$ And trivially $$p\mid p(2k(10n+4)+k^2p)$$