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Conjecture re 32n^2 + 3n

  1. May 8, 2010 #1
    Conjecture: If P is odd, then there is one and only one number n in the set {1,2,3,...(P-1)} which satisfies the equation (32*n^2 + 3n) = 0 mod P an this number. Can anyone help me with a proof of this? If by chance this is a trival matter. I have gone further and determined 4 equations for n based upon the value of P mod 8, but I will leave that for later. I would like to know if the conjecture is trival first.
  2. jcsd
  3. May 9, 2010 #2
    This is just a linear equation since we can factor as (n)(32n+3), and n==0 is unacceptable.

    So we are left with [tex]32n+3\equiv 0 \bmod p [/tex]
  4. May 9, 2010 #3
    Question doesn't say P is prime, just odd, so shouldn't this be fleshed out a bit?
  5. May 9, 2010 #4


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    I think that's sufficient, since gcd(P, 32) = 1.
  6. May 9, 2010 #5
    I wasn't talking about getting a solution of 32n+3=0(P). I thought the reasoning behind

    [itex]n(32n+3)=0(P)\Rightarrow n=0(P)\vee 32n+3=0(P)[/itex]

    was a bit sparse. This would be obvious if P were prime, but what's wrong with the following?


    Here 9 and 18 are different residues modulo 27, neither 0(27). Also 32n+3=0(27) is false for both n=9 and n=18.
    Last edited: May 9, 2010
  7. May 10, 2010 #6
    I was just thinking of primes, but, assumedly, your reasoning is correct.
  8. May 11, 2010 #7
    I am not sure if I got the question right but with P=9, 32n^2+3n=0 (mod 9) for n=3 and n=6.
  9. May 11, 2010 #8
    P=55; n=11,25
  10. May 12, 2010 #9
    Thanks everyone for the counter examples. The ones with 3|n should have been obvious.
    Last edited: May 12, 2010
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