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.

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

Exactly.

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.