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…


Source :- RMO 2015 West Bengal Problem 3

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


Let us put \(x=0.\) We get \(-\)


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.

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