Prove that there exist integers $x, y$ so that $2x^2+3y^2-r$ is divisible by some fixed prime larger than 3.
$r$ is also an integer.
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1To clarify : For any given $r$, such $x$ and $y$ have to be found , right ? – Peter Feb 21 '21 at 19:44
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Hint: there are $\frac {p+1}2$ residues of the form $2n^2 \pmod p$ and the same number of the form $-3m^2+r\pmod p$. (at least for $p>3$) – lulu Feb 21 '21 at 19:52
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See. e.g., this question for a similar problem. – lulu Feb 21 '21 at 19:52
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We are given $p, r$. – gournge Feb 22 '21 at 06:11
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Let $p > 3$ be the fixed prime.
We consider the following $p + 1$ numbers:
- $2x^2$ for $x = 0, 1, \dots, \frac {p - 1}2$;
- $r - 3y^2$ for $y = 0, 1, \dots, \frac {p - 1}2$.
Since there are only $p$ different residues mod $p$, there must be two numbers with the same residue.
It is clear that
- $2x_1^2 \neq 2x_2^2$ for any $0 \leq x_1 < x_2 \leq \frac{p - 1}2$;
- $r - 3y_1^2 \neq r - 3y_2^2$ for any $0 \leq y_1 < y_2 \leq \frac{p - 1}2$.
Therefore there must exist $x, y$ such that $2x^2 = r - 3y^2$.
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