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December 7, 2015

RMO 2015 Problem 3 | West Bengal Region

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...


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 comments on “RMO 2015 Problem 3 | West Bengal Region”

  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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