# Not a perfect square of the form for any integer x.

Now a days, I become good fan of this site, as this site making me to learn more math..hahaha. Okay!

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, ...

• 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!) – louie mcconnell Sep 23 '14 at 4:42
• $$\sum_{n=1}^x 2n-1.$$ @louiemcconnell just for your $\LaTeX$ education, right click and show math as latex. Wrap that mess in dollars. :) – J. W. Perry Sep 23 '14 at 4:47
• 2 cool features, 1 comment. Thanks! – louie mcconnell Sep 23 '14 at 4:49
• @louiemcconnell Being mathematically proper, I should have wrapped that $2n-1$ in parentheses to properly represent your idea. – J. W. Perry Sep 23 '14 at 5:00

## 2 Answers

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.

• Very well spotted! – Jyrki Lahtonen Sep 23 '14 at 5:34
• @Andre Nicolas! can you prove that, why 4k+3 does not divide y^2 + 1? – user177889 Sep 23 '14 at 11:48
• 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) – André Nicolas Sep 23 '14 at 15:27
• (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. – André Nicolas Sep 23 '14 at 15:33
• 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? – Kieren MacMillan Oct 29 '14 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.