Conjecture: Solving 32n^2 + 3n = 0 mod P for odd P

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Discussion Overview

The discussion revolves around the conjecture that for an odd integer P, there exists a unique n in the set {1, 2, 3, ..., (P-1)} such that the equation (32n^2 + 3n) ≡ 0 (mod P) holds. Participants explore the implications of this conjecture, particularly in relation to whether P must be prime and the nature of solutions in different modular contexts.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant conjectures that there is a unique solution n for the equation (32n^2 + 3n) ≡ 0 (mod P) when P is odd, seeking proof or clarification on the conjecture's triviality.
  • Another participant points out that the equation can be factored into (n)(32n + 3) and notes that n = 0 is not acceptable, leading to the simplified equation 32n + 3 ≡ 0 (mod P).
  • Some participants question whether P being merely odd is sufficient, suggesting that the nature of P (whether prime or not) should be clarified.
  • Concerns are raised about the reasoning behind the factorization, especially in cases where P is not prime, with examples provided to illustrate potential issues with multiple residues yielding the same modular result.
  • Counterexamples are presented where specific values of n satisfy the equation for composite values of P, such as P = 9 and P = 55, indicating that the conjecture may not hold universally.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the conjecture, with some providing counterexamples that challenge the uniqueness claim. The discussion remains unresolved regarding the conditions under which the conjecture holds true.

Contextual Notes

Participants highlight the importance of distinguishing between odd and prime values of P, as well as the implications of the gcd condition with respect to the coefficients in the equation. There are unresolved questions about the nature of solutions in non-prime modular arithmetic.

ramsey2879
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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.
 
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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]
 
robert Ihnot said:
This is just a linear equation since we can factor as (n)(32n+3), and n==0 is unacceptable.

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

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

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:
Martin Rattigan said:
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?

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.

I was just thinking of primes, but, assumedly, your reasoning is correct.
 
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.
 
P=55; n=11,25
 
Martin Rattigan said:
P=55; n=11,25
Thanks everyone for the counter examples. The ones with 3|n should have been obvious.
 
Last edited:

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