number theory, prove that $10m+n\equiv 0 \pmod {11}$ suppose $m, n$ are two nonzero integers, if $m^2+n^2$ can be divided by $mn$, prove that $10m+n\equiv 0 \pmod {11}$. 
I have tried some ways,
$m^2+n^2=kmn$, so $m|n^2$ and $n|m^2$, but I can't make a connection to number $11$, thanks for help !
 A: Let $d=\gcd(m,n)$
Then $m=dq$ and $n=dj$ where $\gcd(q,j)=1$
It is given that $\frac{d^2(q^2+j^2)}{d^2.qj}$ must be an integer.
Then 
$\frac{q^2+j^2}{qj}$ must be an integer too .
Hence $\frac{q}{j} + \frac{j}{q} $ must be integral. Multiplying by q  we get that $j+ \frac{q^2}{j}$ must be integral too.
From here you can get $q|j^2$ and $ j|q^2$.But we know that $\gcd(q,j)=1$ , don't we?
Hence we must have $q=j=1$.
Hence $m=n$. 
Hence $10m+n=11m$ or $11n$ ( whichever you like) .
Thus, the claim is proved
A: We have $m^2+n^2-kmn=0$ for some integer $k$.
For $k$ fixed, let's denote by $M,N$ the principal values (between $0$ and $10$) of the roots of the previous equation taken mod $11$ which minimizes the sum $m+n$ and such that $M\neq M$.
We can assume WLOG by symmetry that $M> N\geq0$
Let's consider the equation $x^2+N^2-kNx=0=x^2-sx+p$ where $s,p$ are the sum and the product of the roots respectively. We know that $x_1=M$ is one root.
We have $p=x_1x_2=N^2=Mx_2\rightarrow x_2=\frac{N^2}{M}$ $(M>0)$
Since $M,N$ are roots such that the sum is minimized, we have $M+\frac{N^2}{M}\geq M+N \rightarrow N\geq M$ (assuming $N\neq 0$), which is a contradiction.
If $N=0$, then $M=0$ and $N=M$ which is also a contradiction.
Therefore, there are no roots whose distinct principal values minimize the sum $m+n$, which means the roots are necessarily equal (principal-value-wise, mod $11$)
Hence, $m\equiv n \pmod {11}$
A: Like Avi's answer, I will prove $m=n$, using quadratic equation.
Since $ mn | m^2+n^2$, there exists a positive integer $k$ such that $kmn=m^2+n^2$. In other words, 
     $$x^2-knx + n^2=0 $$ 
has an integer solution. 
This means $$\Delta=(k^2-4)n^2$$ is a square, so $k^2-4$ is a square. This is only possible when $k=2$, as when $k\ge 3$, we have $(k-1)^2<k^2-4<k^2$.
When $k=2$, we have $m=n$.
(Remark: I assumed that $m>0$ and $n>0$, so $k$ must be positive. If not, it is possible that $m=-n$ and the OP's statement is not true.)
A: I don't know if it's kosher to post half an answer outside the comments, but in order to get more attention:  I think the given solution is missing something cute, but I can't work it out. Note:
$$(10m+n)^2 = 100m^2+20mn+n^2 \equiv 99m^2+m^2+22mn-2mn+n^2 = (m-n)^2 \pmod{11}.$$
I don't know why I think this gives more information that just observing
$$10m+n \equiv -m+n \pmod{11}$$
but I liked the $99$ and $22$ didn't want to lose it.  It sure seems like this observation should have a cute punchline....?
