Integer solutions of $x^3-x+9=5y^2$

Question

What are the solutions in integers of $x^3-x+9=5y^2$?

[Source: Hungarian competition problem]

Answer 1

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.

Answer 2

This is the best I could do so far:

$$x^3-x+9=5y^2$$

Let $$x=247$$

Then

$$247^3-273+9=5y^2\implies y^2=3013797 $$

Which gives $$y=1736.02909\approx 1736$$

This is the closest I've got so far to an integer for y.

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The problem is even such small decimals give to large difference (in this case it's only 5). Im sharing this because it mighty help maybe also there is no easy solution (no perfect solution too). But after 1 hours of work this is the nearest I have seen.

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Also, if you didnt know

$$y^2=\frac{x^3-x+9}{5}$$

$y^2$ is a integer

when $x$ is 7,17,27,37,.........,1127,.........,18887,etc

(means when the value x has a 7 as the last digit)

So this will allow you to narrow your Search.