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

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.

  • $\begingroup$ Or $x=8,y=12$ or $x=23,y=33$. $\endgroup$ – lulu Jan 2 at 20:08
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    $\begingroup$ In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions. $\endgroup$ – Sameer Kailasa Jan 2 at 20:08
  • $\begingroup$ Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean? $\endgroup$ – lulu Jan 2 at 20:10
  • $\begingroup$ 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! $\endgroup$ – 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.


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