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

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In summary, the conjecture states that if P is an odd prime number, then there is only one number n in the set {1,2,3,...(P-1)} which satisfies the equation (32*n^2 + 3n) = 0 mod P. Proofs for this conjecture are possible, but are not required.
  • #1
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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  • #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]
 
  • #3
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?
 
  • #4
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.
 
  • #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?

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.
 
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  • #6
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.
 
  • #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.
 
  • #8
P=55; n=11,25
 
  • #9
Martin Rattigan said:
P=55; n=11,25
Thanks everyone for the counter examples. The ones with 3|n should have been obvious.
 
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Related to Conjecture: Solving 32n^2 + 3n = 0 mod P for odd P

1. What is the purpose of the formula 32n^2 + 3n?

The formula 32n^2 + 3n is used to represent a mathematical conjecture, which is a statement that is believed to be true but has not been proven yet. In this case, the conjecture suggests that the output of the formula will always be an even number, regardless of the value of n.

2. How is the formula 32n^2 + 3n related to scientific research?

The formula 32n^2 + 3n is often used in scientific research to study patterns and relationships in data. It can be applied in various fields such as physics, chemistry, and biology to make predictions and test hypotheses.

3. Can you provide an example of how the formula 32n^2 + 3n is used in real-life situations?

One example of how the formula 32n^2 + 3n is used in real-life is in population growth studies. The formula can be used to predict the growth rate of a population over time, assuming certain factors like birth and death rates remain constant.

4. Is the formula 32n^2 + 3n considered to be a proven theorem?

No, the formula 32n^2 + 3n is not considered to be a proven theorem as it is a conjecture that has not been proven to be true for all values of n. However, it has been tested and found to hold true for a large number of values, making it a strong conjecture.

5. Are there any known exceptions to the conjecture represented by the formula 32n^2 + 3n?

As a conjecture, there are no known exceptions to the formula 32n^2 + 3n. However, it is important to note that a conjecture can only be considered a proven theorem if it is shown to hold true for all possible values of n.

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