TITLE: How prove this equation have infinite solution?
QUESTION [12 upvotes]: Let $x,y,z\in Z$, such that $\gcd(x,y)=\gcd(y,z)=\gcd(x,z)=1$.
Show that the number of solutions to
  $$2013x^2+y^3=z^4$$
  is inﬁnite.

This problem is from the China Mathematical Olympiad (No solution), and I look at it occasionally, but can't prove it.
I also found this harder problem, I think this problem have nice methods 
http://math.univ-lyon1.fr/~roblot/ihp/Fermatlectures.pdf

REPLY [7 votes]: To continue achille hui's discussion, I will prove that if $u=1,v=2n,$ then $(y,z)=1$ and hence $(x,z)=(x,y)=1.$

$x(n)=24 n \left(24156 n^2+1\right) \left(64834704 n^4-5368 n^2+1\right) \left(5251611024 n^4-48312 n^2+1\right)$
$y(n)=\left(24156 n^2-1\right)^4-2013 \left(193248 n^3+8 n\right)^2$
$z(n)=\left(24156 n^2-1\right) \left(583512336 n^4+144936 n^2+1\right)$

Let $s(n)=78774165360 n^4+16087896 n^2-705$
$t(n)=1902868738436160 n^6-740477154384 n^4-12826836 n^2-1343$
We can verify that $y(n)s(n)-z(n)t(n)=-2048$ for all $n\in\mathbb N.$
Hence if $p\mid GCD(y(n),z(n))$ then $p\mid 2048$ hence $p=2.$
However, $y(n)$ and $z(n)$ are both odd for all $n\in\mathbb N.$ Hence $GCD(y(n),z(n))=1$ for all $n\in\mathbb N.$