# Show that there are infinitely many integer solutions for $x^2=y^3+z^5$

Show that there are infinitely many integer solutions for $$x^2=y^3+z^5$$

I can't solve this problem, i did try solving by parametric method of diophantine equations, written each variable in function of a new variable n , trying a form of envolving the variables, example, I suppose that y=xk, such that k is integer..

however, I know that the solutions are the way:

Equation, $$x^2=y^3+z^5$$

$$z^5=(x^2-y^3)$$ -----$$(1)$$

Take $$x=n^{p}(n+1)^{q}$$ &

$$y=n^{r}(n+1)^{s}$$

In equation $$(1)$$, Since exponent of $$'x '$$ is two & exponent of $$'y '$$ is three we take

$$(p+q)=3k$$ & $$(r+s)=2k$$

Hence we get;

$$(q,s)=[(3k-p),(2k-r)]$$

hence we get after substitution for $$(x,y)$$,

$$z^5=x^2-y^3$$

=$$n^{(2p)}(n+1)^{(6k-3r)}$$*$$[(n+1)^{(3r-2p)}-n^{(3r-2p)}]$$

In-order to remove the box bracket above we take exponent,

$${(3r-2p)=1}$$

$$r=(2p+1)/3$$

So we get:

$$z^5=n^{(2p)}(n+1)^{(6k-3r)}$$

Since $$'z'$$ is a fifth power we take exponents, $$(2p)=5*4$$ & $$(6k-3r)=5*3$$

Since, $$p=10$$, then because, $$r=(2p+1)/3$$,

we get, $$r=7$$ & $$k=6$$

Also as, $$q=(3k-p)$$ we get $$q=8$$ and

as, $$s=(2k-r)$$ we get, $$s=5$$

Since, $$(p,q,r,s)=(10,8,7,5)$$ we get,

$$x=n^{10}(n+1)^{8}$$

$$y=n^{7}(n+1)^{5}$$

$$z=n^{4}(n+1)^{3}$$

It's important to note that you are not asked to classify all solutions, just to find an infinite family of them.

Here's a simple, infinite family:

If $$y=2^{5a}$$, $$z=2^{3a}$$ then the left hand is $$2^{15a+1}$$ so you just need to choose $$a$$ to be odd in order to get a square.

For example, with $$a=1$$ we get $$(x,y,z)=(2^8,2^5,2^3)$$

Note that if $$(x,y,z)$$ is a solution then $$(k^{15}x, k^{10}y, k^6z)$$ is also a solution.

Since $$3^2=2^3+1^5$$, we find that $$(k^{15}\cdot 3, k^{10}\cdot 2, k^6)$$ is a solution for all integers $$k$$.