<s>[INST] Prove that series $(10n+4)^2+1$ contains infinitly many composite numbers 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?
 [/INST] 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 $$p\mid p(2k(10n+4)+k^2p)$$
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