Home > IMO, Number theory, Olympiad > Romanian TST II 2011 Problem 4

Romanian TST II 2011 Problem 4


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(a) There are infinitely many positive integers n such that there exists a square equal to the sum of the squares of n consecutive positive integers (for instance, 2 and 11 are such, since 5^2=3^2+4^2, and 77^2=18^2+19^2+...+28^2).

(b) If n is a positive integer which is not a perfect square, and if x_0 is an integer number such that x_0^2+(x_0+1)^2+...+(x_0+n-1)^2 is a perfect square, then there are infinitely many positive integers x such that x^2+(x+1)^2+...+(x+n-1)^2 is a perfect square.

Romanian TST 2011

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Categories: IMO, Number theory, Olympiad Tags: ,
  1. abconjecture
    May 31, 2011 at 5:07 pm

    So these are all the tst problems? Wow they look very nice. Do you have solutions for these?

    • June 1, 2011 at 12:27 am

      I haven’t thought very much on how to solve them, since I had a busy period with exams, but I’ll post solutions as soon as I have time. And I think there are more problems, but these are the ones I have until now.

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