$a^2+b^3=c^5$Are there infinitely many solutions? I am having troubles figuring whether there are infinitely many integer solutions to the following equation: $$a^2+b^3=c^5$$ 
This is just a problem I thought of on my own, so sorry in advance if this is already an open problem. 
The way I tried to solve it is this: 
We know that there are infinitely many integers such that $a^2+y^2=z^2$ . Therefore, if $b^3=y^2$ and $c^5=z^2$ , we might have a solution. This reduces to $y=b^{3/2}$ and $z=c^{5/2}$ . So $b$ and $c$ have to be squares. However, I don't know where to go from here. I don't know how to prove that there are infinitely many $y, z$ which satisfy $y=b^{3/2}$ and $z=c^{5/2}$ $and$ $a^2+y^2=z^2$ . Thanks.
 A: For solutions without common factor:
Beukers found polynomials such that $a=P(t), b=Q(t), c=R(t)$ is a primitive solution for all integer $t$, and Jonny Edwards gave a list of such polynomial curves that include all primitive solutions.
http://igitur-archive.library.uu.nl/dissertations/2006-0208-200155/appd.pdf
Solutions with common factors can be written down by elementary means such as selecting $n$ so that the three terms are equal to $2^n, 2^n$ and $2^{n+1}$.
A: You can also get a solution from $a^2+b^3=d$, by multiplying by $d^{24}$:
$$(ad^{12})^2+(bd^8)^3=(d^5)^5$$
A: We will cheat. Let $a=2^{3k}$ and let $b=2^{2k}$. Then the left-hand side is equal to $2^{6k+1}$.
It is easy to find infinitely many $k$ such that $6k+1$ is divisible by $5$. 
Just let $k$ be any integer congruent to $4$ or $9$ modulo $10$. 
Then we can let $c=2^{(6k+1)/5}$.
Remark: We get into interesting territory when we ask that the numbers be relatively prime. For some information, see the Wikipedia article on Beal's Conjecture.
