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

  • $\begingroup$ is this the original problem?can you give us a link? $\endgroup$ – Konstantinos Gaitanas Oct 13 '13 at 21:03

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.

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  • $\begingroup$ I have some idea: maybe use pell equation $m^2-2013n^2=1$ have Infinitely many integer solution? and let $m=z^2$ $\endgroup$ – china math Oct 12 '13 at 13:58
  • $\begingroup$ May I ask how you found out these convenient polynomials ? They look like norms in some number field $\endgroup$ – Ewan Delanoy Oct 12 '13 at 13:59
  • $\begingroup$ @EwanDelanoy Let $t = z^2$, one can rewrite the equation as $$y^3 = t^2 - 2013 x^2 = (t + x\sqrt{2013})(t - x\sqrt{2013})$$ An ansatz to solve this is to pick $a, b$ from $\mathbb{Z}$ and set $$t + x\sqrt{2013} = (a + b\sqrt{2013})^3\quad\text{ and }\quad y = (a^2 - 2013 b^2)$$ You then ask for under what condition that $t$ is a square. If one repeat above types of ansatz several times, one will ultimately get the 3 polynomials I got. $\endgroup$ – achille hui Oct 12 '13 at 14:30

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 $n\in\mathbb N.$

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  • $\begingroup$ Sorry, where does "hence $(x,z) = (x,y) = 1$ come from? $\endgroup$ – user27126 Oct 13 '13 at 9:00
  • $\begingroup$ @Sanchez For example, if $p\mid x,p\mid z$ then $p\mid 2013x^2-z^4=y^3$ hence $p\mid y$, then $p\mid (y,z),$ a contradiction. $\endgroup$ – lsr314 Oct 13 '13 at 10:09
  • $\begingroup$ so,is the problem solved?why isn't this an accepted answer? But i am really wondering:This is the answer that everybody expected to be found? $\endgroup$ – Konstantinos Gaitanas Oct 13 '13 at 13:29
  • $\begingroup$ @KonstantinosGaitanas, Yes, the problem has been solved. No, since this is an olympiad type of question, one should expect an answer simpler. In particular, there should be a way to answer this question w/o involving polynomials with coefficients that big. The bounty is still open for someone to grab ;-) $\endgroup$ – achille hui Oct 13 '13 at 16:39

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