<s>[INST] Integer Solution to $x^3+y^2=z^2$ What is the non-zero integer general solution to $x^3+y^2=z^2$ ? I guess it is already solved in some book or paper, in that case plz help me to find that.
Edit:


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*One solution is - 
$n^3 = [(n)(n+1)/2]^2 - [(n)(n-1)/2]^2$
Found in  here.

*$(x,y,z)=\left(abuv,\frac{ab(bu^3-av^3)}{2},\frac{ab(bu^3+av^3)}{2}\right)$ where $bu^3\equiv av^3\pmod{2}$
 [/INST] There are a lot of solutions. For example, parameterize a set of such solutions. 
We have modulo $7$ the only cubes are $0$ and $\pm1$ and besides $t^6=1$ for all $t$ non divisible by $7$. Take for example the cube $1$ so we have
$$1\equiv(z-y)(z+y)\pmod7$$ We can choose for example $z+y=t^5$ and $z-y=t$ so we get 
$$z=\frac{t^5+t}{2}\hspace{15mm}y=\frac{t^5-t}{2}$$ This allow us to get the parameterization in integers
$$\begin{cases}x=t^2\\y=\dfrac{t^5-t}{2}\\z=\dfrac{t^5+t}{2}\end{cases}$$in relation with the identity $$(t^2)^3+\left(\frac{t^5-t}{2}\right)^2=\left(\frac{t^5+t}{2}\right)^2$$
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