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:


*

*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 $bu^3\equiv av^3\pmod{2}$
 A: Let $(x,y,z)\in\mathbb{Z}^3$ be such that $x^3+y^2=z^2$.  Then,
$$x^3=z^2-y^2=(z-y)(z+y)\,.$$
Write
$$z-y=s\,\prod_{i=1}^l\,p_i^{3t_i}\,\prod_{j=1}^m\,q_j^{3u_j+1}\,\prod_{k=1}^n\,r_k^{3v_k+2}\,,$$
where 


*

*$s\in\{-1,+1\}$, 

*$p_1,p_2,\ldots,p_l$, $q_1,q_2,\ldots,q_m$, and $r_1,r_2,\ldots,r_n$ are pairwise distinct prime natural numbers, and 

*$t_1,t_2,\ldots,t_l$, $u_1,u_2,\ldots,u_m$, and $v_1,v_2,\ldots,v_k$ are nonnegative integers.


Take
$$a:=\prod_{j=1}^m\,q_j\,,$$ $$b:=\prod_{k=1}^n\,r_k\,,$$ and $$c:=s\,\prod_{i=1}^l\,p_i^{t_i}\,\prod_{j=1}^m\,q_j^{u_j}\,\prod_{k=1}^n\,r_k^{v_k}\,.$$
Then, $z-y=ab^2c^3$ with $a$ and $b$ being squarefree positive integers, and $c$ being an integer.  Because $x^3=(z-y)(z+y)$, it follows immediately that
$$z+y=a^2bd^3\,,$$
for some integer $d$.  Consequently,
$$x=\sqrt[3]{(z-y)(z+y)}=abcd\,,\tag{*}$$
$$y=\frac{z+y}{2}-\frac{z-y}{2}=\frac{ab(ad^3-bc^3)}{2}\,,\tag{#}$$
and
$$z=\frac{z+y}{2}+\frac{z-y}{2}=\frac{ab(ad^3+bc^3)}{2}\,.\tag{$\star$}$$
For $y$ and $z$ to be integers, we need that
$$ab(c-d)\equiv ab(ad^3\pm bc^3)\equiv 0\pmod{2}\,.$$
Thus, a solution $(x,y,z)$ is uniquely determined by 


*

*two squarefree positive integers $a$ and $b$ which are relatively prime, and

*two integers $c$ and $d$ such that $ab(c-d)$ is divisible by $2$
via the formulae (*), (#), and ($\star$) above.  If we relax the first condition by allowing $a$ and $b$ to be arbitrary integers, then more than one quadruples $(a,b,c,d)$ may produce the same triple $(x,y,z)$.
Here are some infinite families presented by other users.  The infinite family $$(x,y,z)=\left(N,\dfrac{N(N-1)}{2},\dfrac{N(N+1)}{2}\right)$$ for $N\in \mathbb{Z}$ corresponds to $(a,b,c,d)=(N,1,1,1)$.  The infinite family $$(x,y,z)=\left(N^2,\dfrac{N(N^4-1)}{2},\dfrac{N(N^4+1)}{2}\right)$$ for $N\in \mathbb{Z}$ corresponds to $(a,b,c,d)=(N,1,1,N)$.
A: Equation $(x^3+y^2=z^2)$ has parametric solution:
$x=(2k-1)$
$y=(k-1)(2k-1)$
$z=k(2k-1)$
For, $k=7$ we get:
$(x,y,z)=(13,78,91)$
A: 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 $$(t^2)^3+\left(\frac{t^5-t}{2}\right)^2=\left(\frac{t^5+t}{2}\right)^2$$
