I am preparing myself for my number theory midterm and I am stuck in this textbook question. I don't understand what question is asking. Any hint would be appreciated.

Prove that if $x$ is an integer then $x^3 + y^2 = z^2$ has a solution in integers $y$, $z$.

My attempt: $x^3 = z^2-y^2 = (z-y)(z+y)$, what is the next step?


See the point is we need to find integers $y$ and $z$ such that $z^2-y^2 = x^3$. To do this, we factorize : $(z-y)(z+y) = x^3$. We will be done if there exist $y,z$ such that $z + y = x^2$ and $z-y = x$. Fortunately, $$ y= \frac{x^2-x}{2}, z = \frac{x^2+x}{2} $$ satisfy these conditions, and furthermore are integers, because if $x$ is even, then $x^2-x$ and $x^2+x$ are even, and if $x$ is odd, then $x^2$ is odd, so $x^2-x$ and $x^2+x$ are even. Hence, $y$ and $z$ satisfy $x^3+y^2 = z^2$.

To check the above for a large value, let $x=7$, and note that $x^2 = 49$, so $z = 28$ and $y = 21$, so that $7^3 + 21^2 = 441 + 343 = 784 = 28^2$.


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