This is a Regional Mathematics Olympiad, RMO 2015 Problem 3 from West Bengal Region.
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.
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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.