# How to solve the following diophantine equation?

$$2x^3 + x + 8 = y^2$$

WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $$LHS$$ can never be a perfect square.

• Or $x=8,y=12$ or $x=23,y=33$. – lulu Jan 2 at 20:08
• In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions. – Sameer Kailasa Jan 2 at 20:08
• Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean? – lulu Jan 2 at 20:10
• Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer! – Krisztián Kiss Jan 2 at 20:20

The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached WA link refers to the equation $$2x^3+x+8=y^2$$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $$\pmod 3$$. Since $$n^3\equiv n\pmod 3$$ for all $$n$$, we see that the LH is always $$2\pmod 3$$. But $$2$$ isn't a quadratic residue $$\pmod 3$$ so we are done.