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Can we prove that $x^3 + 7$ cannot be perfect square for any positive/negative or odd/even integer of $x$.

I checked with number up to x = 1,...1000. I realized that, not a perfect. But, how do we prove the statement without further checking in place of 1001, ...

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    $\begingroup$ Just an idea. Maybe something along the lines that any perfect square can be alternatively represented as a summation of all numbers from n=1 to x for 2n - 1 (please help, latex pros!) $\endgroup$ Commented Sep 23, 2014 at 4:42
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    $\begingroup$ $$\sum_{n=1}^x 2n-1.$$ @louiemcconnell just for your $\LaTeX$ education, right click and show math as latex. Wrap that mess in dollars. :) $\endgroup$ Commented Sep 23, 2014 at 4:47
  • $\begingroup$ 2 cool features, 1 comment. Thanks! $\endgroup$ Commented Sep 23, 2014 at 4:49
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    $\begingroup$ @louiemcconnell Being mathematically proper, I should have wrapped that $2n-1$ in parentheses to properly represent your idea. $\endgroup$ Commented Sep 23, 2014 at 5:00

2 Answers 2


Rewrite our Diophantine equation as $y^2+1=x^3+8$. Note that $x$ cannot be even, for $y^2+1$ cannot be divisible by $4$.

So $x$ must be odd. We have $$y^2+1=x^3+8=(x+2)(x^2-2x+4).$$ Suppose $x\equiv 1\pmod{4}$. Then $x+2\equiv 3\pmod{4}$, which is impossible, since a positive number of the form $4k+3$ cannot divide $y^2+1$.

Suppose $x\equiv 3\pmod{4}$. Then $x^2-2x+4\equiv 1-2(3)\equiv 3\pmod{4}$, again impossible.

  • $\begingroup$ Very well spotted! $\endgroup$ Commented Sep 23, 2014 at 5:34
  • $\begingroup$ @Andre Nicolas! can you prove that, why 4k+3 does not divide y^2 + 1? $\endgroup$
    – user177889
    Commented Sep 23, 2014 at 11:48
  • $\begingroup$ If a number is of the form $4k+3$, then it has a prime divisor $p$ of the form $4k+3$ (because a product of powers of primes of the form $4k+1$ is also of form $4k+1$.) So let $p$ be a prime divisor of $y^2+1$ of the form $4k+3$. Then $y^2\equiv -1\pmod{p}$, contradicting the fact that $-1$ is never a quadratic residue of a prime of the form $4k+3$. How do we prove that $y^2\equiv -1\pmod{p}$ has no solution if $p\equiv 3\pmod{4}$. Well, it is early in every elementary number theory book. It has also been proved several times on MSE, by me and bunches of others.(Cont) $\endgroup$ Commented Sep 23, 2014 at 15:27
  • $\begingroup$ (Cont) Quickest proof goes like this: if $y^2\equiv -1\pmod{p}$, then $y$ has order $4$ modulo $p$, and therefore $4$ divides $p-1$, i.e. $p$ has form $4k+1$. Sorry not to provide MSE links, but I am lousy at searching, specially Before Coffee. $\endgroup$ Commented Sep 23, 2014 at 15:33
  • $\begingroup$ Don't you need to show that $\gcd(x+2,x^2-2x+4)=1$, to eliminate the possibility of that hypothetical prime factor $4k+3$ being squared? I know it's trivial to show, but I think you should/must show it, no? $\endgroup$ Commented Oct 29, 2014 at 15:39

This is a Mordell equation: $$ x^3+7=y^2 $$ and all Mordell equations have finitely many solutions. You can see at A054504 in the OEIS that this particular equation has no solutions, but I don't know of an easy proof. With this sort of Diophantine equation there may be no easy proof.


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