<s>[INST] Integer solutions of $x^3-x+9=5y^2$ What are the solutions in integers of $x^3-x+9=5y^2$?
[Source: Hungarian competition problem]
 [/INST] If $x$ were odd, then we would have $x^3 \equiv x \pmod {8}$.  But $5y^2 \equiv 1\pmod{8}$ has no solutions, so $x$ is even.
We have $(x-1)x(x+1) = 5y^2 - 9 \equiv 1\pmod{5}$.  We have $2\cdot 3\cdot 4 \equiv -1\pmod{5}$, so we must have $x\equiv 2\pmod{5}$.
Finally, we observe that $x^3-x$ is divisible by $3$, so $y$ is divisible by $3$, so $x^3-x$ is divisible by $9$.  However, $5y^2 \equiv 9 \pmod{27}$ has no solutions, so $x^3-x$ is divisible by $9$, but not $27$.  This implies that either $x+1$ is not divisible by $3$, or it is divisible by $9$, but not $27$.
In the former case, we have $x+1 \equiv 3\pmod{5}$.  In the latter, $\frac{x+1}{9} \equiv 2\pmod{5}$.  Since $x+1$ is odd, in both cases there is a factor of $x+1$ that is in $\{2,3\}\pmod{5}$, but is not divisible by $2$ or $3$.  That implies that there exists a prime factor $p\neq 2,3$ of $x+1$ such that $p\in \{2,3\}\pmod{5}$.
So $p$ is not a square$\pmod{5}$.  By quadratic reciprocity, $5$ is not a square$\pmod{p}$.  But we have $5y^2 - 9 \equiv 0 \pmod{p}$, so $5\equiv (3/y)^2\pmod{p}$, contradiction.
We conclude that there are no solutions.
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