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 !

  • $\begingroup$ The result is true if $n=m$ ( check by induction) $\endgroup$ – Maman Sep 28 '16 at 16:16
  • $\begingroup$ If isn't enough to prove it you need iff $\endgroup$ – JukesOnYou Sep 28 '16 at 16:20
  • $\begingroup$ @Maman m=n is indeed the case here :) (check out my answer ). $\endgroup$ – user369582 Sep 28 '16 at 16:25

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

  • $\begingroup$ Why did you add $q\mid j^2$ and $j\mid q^2$ ? $\endgroup$ – Maman Sep 28 '16 at 16:37
  • $\begingroup$ $\frac{q+j}{qj}=\frac{q}{qj}+\frac{j}{qj}=\frac{1}{j}+\frac{1}{q}$ $\endgroup$ – Maman Sep 28 '16 at 16:54
  • $\begingroup$ Ok nice it works now $\endgroup$ – Maman Sep 28 '16 at 17:01
  • 2
    $\begingroup$ Why is $\frac{d^2(q^2+j^2)}{d^2.q^2.j^2}$ an integer ? It is only given that $\frac{d^2(q^2+j^2)}{d^2.q.j}$ is an integer. $\endgroup$ – Evariste Sep 28 '16 at 17:09
  • $\begingroup$ First I multiply by q and deduce that$ j|q^2 $ since $\frac{q^2}{j} $is integral. Then , instead of multiplying by q , I multiply by j and deduce that $q|j^2 $ as in that case $\frac{j^2}{q} $ will be integral. $\endgroup$ – user369582 Sep 28 '16 at 17:57

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, $m\equiv n \pmod {11}$


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.)


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....?


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