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

Question

Prove/disprove that $x^3 - x + 9 = 5 y^2 $ has integer solution. i.e) There doesn't exist $x,y$ in integers satisfying above equation.

I tried using congruence and get if solution exist then $x$ should be congruent to $2\pmod{5}$. But after that, I tried something but I haven't get any conclusion. Please give me a hint if the statement is true.

Answer 1

The substitution $U=5x, V=25y$ turns this equation into that of an elliptic curve with the short Weierstrass form $$ U^3-25U+1125=V^2. $$ According to its LMFDB entry the integer points on this elliptic curve are $(U,V)=(4,\pm33)$

There are no solutions with $V$ a multiple of $25$, so the original equation has no integer solutions.

Answer 2

More of a comment, but there are too many already. Having read all the comments, it seems to me a strong possibility that the equation whoever produced the pdf problem sheet intended to set was $$x^2-x+9=5y^2$$

Put another way, the OP insists his Q has $x^3$, but I strongly suspect the Q whoever set the problem wanted him to solve had $x^2$, so it might be worth giving a quick answer to that. @JyrkiLahtonen has already given an excellent answer to the Q as the OP has given it. I hope it has piqued his curiosity for delights to come ...

So completing the square: $$(x-1/2)^2+35/4=5y^2$$ Multiplying by 4 and rearranging: $$5(2y)^2-(2x-1)^2=35$$

Now $2x-1$ must be divisible by 5, so writing it as $5z$ and $2y$ as $w$ we get $$w^2-5z^2=7$$ which is a standard Pell equation. They are discussed endlessly on this site, and it is not hard to show that it has no solution in integers, and a fortiori no solution with $w$ even and $z$ odd).