Show that for any natural number n between $n^2$ and$(n+1)^2$ there exist 3 distinct natural numbers a, b, c, so that $a^2+b^2$ is divisible by c Show that for any natural number n ,one can find 3 distinct natural numbers a, b, c, between $n^2$ and$(n+1)^2$, so that $a^2+b^2$ is divisible by c. 
It's easy to prove that such three distinct numbers exist, by supposing the contrary and coming to contradiction(i.e."suppose $(n+1)^2-n^2=0$ -->$n=-1$, $-1$ is not a natural number, and so on.."), but how to show divisibility?
(The task is from 1998 St. Petersburg City Mathematical Olympiad)


Surprisingly there seems to be another answer to the stricter reading of the problem.
Let $a=(n^2+n)$, $b=(n^2+n+2)$ and $c=(n^2+1)$
and as with @ARoberts Solution if $n\ge 2$ then $n^2 < c < a < b < (n+1)^2$
we have
$a^2+b^2 = (n^2+n)^2 + (n^2 + n + 2)^2 = 2(n^2+1)(n^2+2n+2)$.
So again $c | a^2 + b^2$
