Help on an old Putnam Problem I came across an old Putnam problem that i was having some difficulty with,
and I was wondering if I could get some assistance of this community.
For a given positive integer $m$, I need to find all the triples $(n,x,y)$ of
positive integers, with $n$ relatively prime to $n$, which statisfy
$$(x^2+y^2)^m = (xy)^n.$$
At first it seems like the best place to start is through some form of
relevant base cases, i.e. starting with $m = 2, n = 3.$ i.e. find
$x$ and $y$ solutions such that
$$(x^2+y^2)^2 = (xy)^3$$
$$\implies x^4 + 2x^2y^2 + y^4 = x^3y^3.$$
I can tell that this holds for $x = 0, y = 0,$ but how would I go about finding
all cases from even this one example? This seems very unweildy when we consider
high degree polynomials. What is the best strategy to tackle this kind of
problem?
 A: John Scholes answers this (and almost all Putnam problems) on his site:
https://mks.mff.cuni.cz/kalva/putnam/psoln/psol923.html
It's rather terse, so expanding on it a little here; start by extracting the $gcd(x,y)=g$
$x=gX, y=gY$  
$((gX)^2+(gY)^2)^m=(gXgY)^n$
$\therefore (g^2X^2+g^2Y^2)^m=g^{2n}(XY)^n$
$\therefore g^{2m}(X^2+Y^2)^m=g^{2n}(XY)^n$
$\therefore (X^2+Y^2)^m=(XY)^ng^{2(n-m)}$  
Now $X$ and $Y$ have no shared divisors, because we divided out the $gcd$. So if $p$ is a divisor of $X$ then it divides the $RHS$ but not the $LHS$, because $p$ would also have to divide $Y$, but $Y$ does not share the divisor $p$. So $X=1$. By the same argument $Y=1$. So that last line becomes:  
$(1 + 1)^m=(1.1)^ng^{2(n-m)}$
$\therefore 2^m=g^{2(n-m)}$  
That same line also tells us that $g | (X^2+Y^2)^m=2^m$, so $g$ is a power of $2$.  
$x=gX = g.1 = 2^r$, for some integer $r$.  Similarly for $y=2^r=x$.  
We now have $x,y$ so substituting into the original equation:  
$(2^{2r}+2^{2r})^m=(2^r.2^r)^n$
$\therefore 2^{(2r+1)m}=2^{2rn}$
$\therefore (2r+1)m=2rn$  
But $2r+1$ and $2r$ are coprime. So all the factors of $2r+1$ must be in $n$, or $n=N(2r+1)$ and similarly $m=M.2r$. Substituting and cancelling gives $M=N$. But $m,n$ are coprime. So $M=N=1$. 
So in summary $x=y=2^r$, $m=2r, n=2r+1$. Or alternatively, $m$ must be even and the only solution is $n=m+1, x=y=2^{m/2}$.
A: Another place to find discussion of old Putnam problems is: The Art of Problem Solving
