*Originally posted by slappy115*

**3n^2+3n+1 will either produce a prime congruent to 1 (mod 6) or the factors of the number produced will all be congruent to 1 (mod 6). I can show that any number that will be produced by it is congruent to 1 (mod 6) but I can't show that factors will be. Contradiction proofs will not work because any number congruent to 5 (mod 6) will also work. Don't reply if you don't know what (mod ðŸ™„ means.**

(6k+5)(6l+1) = 36kl + 30l + 6k +5 = 6m+5

This shows that if only one of the factors of a number is 5 mod 6, then the number itself will be 5 mod 6.

(6k+5)(6l+5) = 36kl + 30k + 30l + 25 = 6m+1

This shows that if two factors are 5 mod 6, the numbe will be 1 mod 6.

From these two lines and what you showed (result is 1 mod 6) we can conclude that IF there are factors that are 5 mod 6, they will come in pairs (even amounts).

That's as far as I could reason in my 5 min break :p