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 🙄 means.
Originally posted by slappy115(6k+5)(6l+1) = 36kl + 30l + 6k +5 = 6m+5
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 🙄 means.
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
Originally posted by slappy115WARNING: Conclusions that have no practical value may follow!
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 🙄 means.
3n^2+3n=6(n+1)(n/2)
3n^2+3n=3n(1+n)
3n^2+3n+1=(6a+1)(6y+1)
n(n+1)=2(a+y+6ay)
a=(91-y)/(1+6y)
Just random conclusions...