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

## Romanian TST II 2011 Problem 4

Show that

(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

Categories: IMO, Number theory, Olympiad Tags: ,
1. 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.