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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.