<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] Since $n^2+2n<(n+1)^2$ and since $n^2$ is included, one obvious answer is
$$\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
Extended exposition
Since $n^2+2n<(n+1)^2=n^2+2n+1$, let $a=(n^2+n)$, $b=(n^2+2n)$ and $c=n^2$ then
$$\frac{a^2+b^2}{c}=\frac{\left(n^2+n \right)^2 +\left(n^2+2n \right)^2 }{n^2}$$
$n^2$ factors out of the numerator and we have
$$\frac{a^2+b^2}{c}=\frac{n^2\left(\left(n+1 \right)^2 +\left(n+2 \right)^2\right) }{n^2}=\left(n+1 \right)^2 +\left(n+2 \right)^2$$
Since we now think $n^2$ is not included this simple answer is precluded and left for reference only. 
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