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

Question

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:

Answer 1

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}$

Answer 2

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 $k$.

Answer 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 $$(x,y,z)=(2^8,2^5,2^3)$$