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Let $x$ and $y$ be integers with $x \neq 0$. Prove that if $x^6+(x^2+y)^3$ is a perfect square, then $y \equiv 0 \pmod{x^2}$.

We can expand the given expression as $$x^6+(x^2+y)^3 = 2x^6+3x^4y+3x^2y^2+y^3 = (2x^2+y)(x^4+x^2y+y^2)=k^2,$$ for some integer $k$. Now assume $x \neq 0$. We proceed by contradiction. Let $y = mx^2+d$, where $0 < d \leq x^2-1$ for some $m$. Then the original equation is equivalent to $$((2+m)x^2+d)(x^4+mx^4+dx^2+(m^2x^4+2dmx^2+d^2))$$ and to $$((2+m)x^2+d)((1+m+m^2)x^4+(d+2dm)x^2+d^2).$$ This seemed like a computational way of solving this, but could we still solve it like this?

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    Also, if $x=0$, then $y$ can be any square. – 2017-01-23
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    Going down a different route leads to the question what the primitive (pairwise coprime) solutions of $r^3 + s^3 + 2t^3 = 0$ are. – 2017-01-23
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    @DanielFischer Do you mean to proving that $y$ is a multiple of $x^2$ here? – 2017-01-23
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    It would lead to that. – 2017-01-23

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As far as I am aware, this is how one can prove it.

Let $a,b$ be integers that satisfy your condition.

So $$a^6+(a^2+b)^3=t^2$$ Then, dividing by $a^6$, we have $$1+\left(1+\frac{b}{a^2} \right)^3=\left(\frac{t}{a^3} \right)^2$$ However, the elliptic curve $1+x^3=y^2$ is an elliptic curve with rank $0$, , so it only has finitely many rational points, all of which are “torsion”, or of finite order. As it has rational points of finite order it follows from the the Nagell-Lutz theorem, that all rational points are integer points.

As $\left(1+\frac{b}{a^2}, \frac{t}{a^3} \right)$ is a rational point on $1+x^3=y^2$, it follows that $1+\frac{b}{a^2}$ is an integer, or $$b \equiv 0 \pmod{a^2}$$ As desired. We are done. (Also see here).