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\centerline{\bf SOLUTION FOR JANUARY 2016}
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A correct solution was turned in by 
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\centerline{William Liu}

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{\bf Problem from December 2015}
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Show that the only solution of:
$$ m^2 + n^2 + p^2 = 2mnp$$
where $m,n,$ and $p$ are integers is $m=n=p=0$.

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{\bf Proof} Notice that if $m=n=p=0$ then the above equation is satisfied so now let us suppose $(m,n,p)$ is a solution with $m,n$ and $p$ all nonzero and further let us assume that $m,n$ and $p$ do not have a common factor.
It follows that since 
$$ m^2 + n^2 + p^2 = 2mnp \eqno(1)$$ then at least one of $m,n$ and $p$ is even because the right-hand side of (1) is even and if $m,n$ and $p$ were all odd then $m^2$, $n^2$, and $p^2$ would all be odd and their sum would be odd so the left-hand side of (1) would be odd but the right-hand side is even and so we would obtain a contradiction.

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So now let us assume without loss of generality that $p$ is even. So then $p=2q$ for some integer $q$ and substituting into (1)  and rewriting yields
$$ m^2 + n^2 = 4mnq-4q^2=4q(mn-q). \eqno(2) $$

Thus the right-hand side of (2)  is even so then either both $m$ and $n$ are even or both $m$ and $n$ are odd. Now $m$ and $n$ cannot both be even because if so then we would have that $m,n$ and $p$ are all even but we assumed that $m,n$ and $p$ have no common factor. Thus $m$ and $n$ must both be odd. Thus $m= 2m_1+1$ and $n = 2n_1 +1$ for some integers $m_1$ and $n_1$.
Substituting this into (2) and multiplying this out gives
$$ 4m_1^2  + 4m_1 + 4n_1^2 + 4n_1 + 2 = 4q(mn-q). $$
Dividing by 2 gives

$$ 2m_1^2  + 2m_1 + 2n_1^2 + 2n_1 + 1 = 2q(mn-q). $$
Now the left-hand side is odd and the right-hand side is even so we obtain a contradiction.
So it must be that $m,n,$ or $p$ is 0 and then it follows from (1) then the other two must also be 0. This completes the proof. 

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