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Posers and Puzzles

Posers and Puzzles

  1. Standard member slappy115
    Slappy slap slap
    25 Nov '05 08:45
    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.
  2. Standard member TheMaster37
    Kupikupopo!
    25 Nov '05 11:22
    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.
    (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
  3. Standard member slappy115
    Slappy slap slap
    26 Nov '05 22:13
    Thanks for the input. I've tried looking at that before and have come to the conclusion too. That, however, doesn't show the factors are congruent to 1 (mod 6) but shows that it is possible for the factors to be congruent to 5 (mod 6) since both 1 and 5 are relatively prime to 6.
  4. Standard member TheMaster37
    Kupikupopo!
    27 Nov '05 09:41
    Yeah, but now you know the factors 5 mod 6 must come in pairs

    I don't know if that's something usefull though. I'm trying a different method now, I'm changing the formula mod 6, in the hope I can factorize it...so far no luck.
  5. Standard member slappy115
    Slappy slap slap
    27 Nov '05 19:43
    I'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.
  6. 28 Nov '05 12:56
    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.
    WARNING: Conclusions that have no practical value may follow!

    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...
  7. Standard member slappy115
    Slappy slap slap
    28 Nov '05 18:53
    Thanks for your input. I will take a look at these and see if they help me with what I have.