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)

 A: This seems to work with the stricter reading of the problem.
Let $a = n^2 + 2$, $b=n^2+n+1$ and $c=n^2+1$.  
If $n \geq 2$ then $n^2 < c < a < b < (n+1)^2$.  In particular $a,b,c$ are distinct.
Moreover $$a^2+b^2 = (n^2+2)^2 + (n^2 + n + 1)^2 = (2n(n+1)+5)(n^2+1).$$
So $c | a^2 + b^2$.
I found this by looking at triples $(a,b,c)$ with the required property for small values of $n$ and noticing a pattern. There seem to be lots of other triples which also have the required property; I'm not sure if these can be parameterized nicely. 
A: 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 $c | a^2 + b^2$
A: 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. 
