<s>[INST] Show that a following equation has no solution in integers: $x^3-x+9=5y^2$ 
Show that a following equation has no solution in integers: $$x^3-x+9=5y^2$$

Clearly we see that $y$ is odd, so $y^2\equiv_8 1$ and thus $8\mid x^3-x-4$.
So if $x$ is odd, then $x-1$ and $x+1$ one is divisible by $2$ and other by $4$ so $8\mid x^3-x$ and thus $8\mid 4$ which is not true. So $x$ must be also even and even divisible by $4$ but not by $8$.
Also if we take mod 5 we get $x^3\equiv_5 x+1$ and thus $$x^2 \equiv_5 x^6 \equiv_5 (x+1)^2 = x^2+2x+1\implies 2x\equiv_5 -1 \implies x\equiv_5 -3$$
If we look at modulo 3 we have $$x^3-x+9\equiv_3 0\implies 3\mid y \implies 9\mid (x-1)x(x+1)$$ 
so $x \equiv_9 0,\pm1$. 
but I can not go any further.
 [/INST] Aqua, here it is. 
As you've obtained, $x\equiv 2\pmod{5}$. Now, $x(x-1)(x+1)=5y^2-9$. Note that, $x+1\equiv 3\pmod{5}$, and is odd. I now claim that, there is a prime $p\mid x(x-1)(x+1)$ such that $p\equiv 2,3\pmod{5}$, and $p\neq 3$. If $x\equiv 1,2\pmod{3}$, then the object $x$ clearly has such a prime divisor (indeed, if all prime divisors of $x$ are of form $5k\pm 1$ then it cannot be congruent to $2$ in modulo $5$). Similarly, if $x\equiv 0\pmod{3}$, then $x+1$ is not divisible by $3$, and with the same logic, has such a prime divisor. 
Now, isolate such a prime divisor, $p\neq 3$ and $p\equiv \pm 2\pmod{5}$. Observe that, $p\mid 5y^2-9$, that is,
$$
5y^2\equiv 3^2\pmod{p}\Rightarrow (\frac{5}{p})=1,
$$
namely, $5$ is a quadratic residue, modulo $p$. Next, by the quadratic reciprocity  law,
$$
(\frac{5}{p})(\frac{p}{5})=(-1)^{\frac{p-1}{2}\cdot 2}=1\Rightarrow (\frac{p}{5})=1
$$
But since $p\equiv \pm 2\pmod{5}$, this is clearly impossible!
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