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.