How prove this equation have infinite solution?

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

Solution 1:

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 $n\in\mathbb N.$

Solution 2:

This is not an answer because I don't know how to deal with the requirement that $x,y,z$ are pairwise relatively prime. However, the equation

$$2013 x^2 + y^3 = z^4$$

does have infinitely many non-trivial solutions. Define $$\begin{align} x(u,v) = & 12uv (6039v^2+u^2)(4052169v^4-1342u^2v^2+u^4)\\ & \quad\times\;(328225689v^4-12078u^2v^2+u^4)\\ \\ y(u,v) = & (u^2-6039v^2)^4-2013(24156uv^3+4u^3v)^2\\ z(u,v) = & (u^2-6039v^2)(36469521v^4+36234u^2v^2+u^4) \end{align}$$ By brute force, one can verify these 3 functions satisfy $$2013 x(u,v)^2 + y(u,v)^3 = z(u,v)^4$$

Not all $(u,v)$ give us triplet $(x,y,z)$ that are pairwise relative prime. The first pairwise relative prime solution I find is generated by setting $(u,v)$ to $(2,1)$, this give us:

$$(x,y,z) = (192613207049468003592,1321800962978257,-220968344555)$$

Maybe someone will have a better idea how to extract pairwise relative prime solutions from these list of "partial" solutions.