- 25 Nov '05 08:453n^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.
- 25 Nov '05 11:22

(6k+5)(6l+1) = 36kl + 30l + 6k +5 = 6m+5*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 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 - 27 Nov '05 19:43I've also tried a bunch of different methods and ideas to show the factors can't be congruent to 1 (mod 6) and contradictions will not work since the product of two numbers congruent to 5 (mod 6) produce a number congruent to 1 (mod 6). And thanks for the help, I really appreciate it.
- 28 Nov '05 12:56

WARNING: Conclusions that have no practical value may follow!*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 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...