TITLE: $a^2+b^3=c^5$Are there infinitely many solutions?
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: You can also get a solution from $a^2+b^3=d$, by multiplying by $d^{24}$:
$$(ad^{12})^2+(bd^8)^3=(d^5)^5$$