<s>[INST] 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)

 [/INST] 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 $c | a^2 + b^2$
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