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)
On
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.
On
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 $c | a^2 + b^2$
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.