# Conjecture re 32n^2 + 3n

1. May 8, 2010

### ramsey2879

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. May 9, 2010

### robert Ihnot

This is just a linear equation since we can factor as (n)(32n+3), and n==0 is unacceptable.

So we are left with $$32n+3\equiv 0 \bmod p$$

3. May 9, 2010

### Martin Rattigan

Question doesn't say P is prime, just odd, so shouldn't this be fleshed out a bit?

4. May 9, 2010

### CRGreathouse

I think that's sufficient, since gcd(P, 32) = 1.

5. May 9, 2010

### Martin Rattigan

I wasn't talking about getting a solution of 32n+3=0(P). I thought the reasoning behind

$n(32n+3)=0(P)\Rightarrow n=0(P)\vee 32n+3=0(P)$

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

9(32.9+3)=0(27)
18(32.18+3)=0(27)

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
6. May 10, 2010

### robert Ihnot

I was just thinking of primes, but, assumedly, your reasoning is correct.

7. May 11, 2010

### RedGolpe

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.

8. May 11, 2010

### Martin Rattigan

P=55; n=11,25

9. May 12, 2010

### ramsey2879

Thanks everyone for the counter examples. The ones with 3|n should have been obvious.

Last edited: May 12, 2010