12
$\begingroup$

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

$\endgroup$
  • $\begingroup$ is this the original problem?can you give us a link? $\endgroup$ – Konstantinos Gaitanas Oct 13 '13 at 21:03
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
7
+50
$\begingroup$

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 $n\in\mathbb N.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.