MHB How to Prove 10m+n mod 11=0 for Natural Numbers m and n?

[Albert1]

$m,n\in N$ and $\dfrac {m^2+n^2}{m\times n}\in N$
prove :$(10m+n)$ mod $11=0$

 

[johng1]

Spoiler

Given $m\in N\text{ and }n\in N$, if ${m^2+n^2\over mn}\in N$, then m=n:
If $m\neq n$, then I can assume $m$ and $n$ are relatively prime. But then $m$ divides $n^2$, which is impossible.
So $$n\equiv m\pmod{11}$$
$$-m+n\equiv 0\pmod{11}$$
$$10m+n\equiv 0\pmod{11}$$