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Let $p$ be a prime number. Prove that there exists $x∈Z$ for which $p|x^2−x+ 3$ if and only if there exists $y∈Z$ for which $p|y^2−y+ 25$.

I've copied the question verbatim from an elementary number theory assignment. I'm not making much ground on it, but here's what I've determined so far:

Each equation is based on the following sequence:

x^2 - x: 0, 2, 6, 12, 20, 30, 42, 56, 72, ...

I'm not sure what to call this type of sequence, but I'm thinking of it as a sequence with an acceleration of $2$, as in, each number on the sequence is increased by a number which is increased by $2$ each iteration.

The only difference between the two equations is the starting point of the sequence: one starts at $3$ while the other starts at $25$. This is how I have been thinking about the problem but unfortunately I think I'm working in the wrong direction.

If anyone could help me move in the right direction of this problem, I would appreciate it

Alex
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  • as to your level: do you know the solutions to $x^2 \equiv -1 \pmod 5 ; ?$ What about the Legendre symbol? – Will Jagy Mar 07 '18 at 01:55

1 Answers1

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Both equations have no solutions mod $2$. Both equations have solutions mod $3$. So, let's assume $p>3$.

The discriminant of $x^2-x+3$ is $-11$. So, $x^2-x+3=0$ has a solution mod $p$ iff $-11$ is a square mod $p$.

The discriminant of $y^2-y+25$ is $-99$. So, $y^2-y+25=0$ has a solution mod $p$ iff $-99$ is a square mod $p$.

Since $-99 = 3^2 \cdot (-11)$, we have that $-11$ is a square mod $p$ iff $-99$ is a square mod $p$.

lhf
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  • Thank you, it seems so obvious now. My major flaw was refusing to think of the question in terms of the equations being congruent to 0 mod $p$. – Alex Mar 07 '18 at 02:35