# RMO 2015 Problem 3

Actually, I had a typographical mistake in my last post i.e. question paper.

The 3rd post says to find integer solutions, not positive integer solutions.

I consider this problem as the easiest problem in RMO 2015. So let’s discuss the solution…

Problem

Source :- RMO 2015 West Bengal Problem 3

Show that there are infinitely many triplets $$(x,y,z)$$ of integers such that $$x^3+y^4=z^{31}.$$

Solution

Let us put $$x=0.$$ We get $$-$$

$$y^4=z^{31}.$$

Now, let $$k,k’$$ be 2 integers, such that $$k’\geqslant 0.$$

If we put $$y=k^{31k’},~z=k^{4k’},$$ then we have $$-$$

$$k^{31k’\cdot 4}=k^{4k’\cdot 31}.$$

Which is true.

Hence, every triplet of the forms, $$\left(0,k^{31k’},k^{4k’}\right),\left(k^{31k’},0,k^{4k’}\right),$$ where $$k,k’$$ are integers, such that $$k’\geqslant 0$$ is a solution to the equation.

And, as $$k,k’$$ takes infinitely many values,  the equation has infinitely many solutions.

This completes the proof.

## 3 Replies to “RMO 2015 Problem 3”

1. Actually obviously it was asking for non zero triplets so its not that easy. You sould try for non zero integers, then it’s a good question

1. But what to do… go through the other questions, you will understand.
I think, unexpected triviality might make it the deciding factor for cut-offs since many of the candidates could be trolled by this.