MHB Positive Integer Solutions of $(x^2+y^2)^n=(xy)^{2014}$

[anemone]

Find all positive integers $n$ for which the equation $(x^2+y^2)^n=(xy)^{2014}$ has positive integer solutions.

 

[Opalg]

Find all positive integers $n$ for which the equation $(x^2+y^2)^n=(xy)^{2014}$ has positive integer solutions.

Spoiler: Partial solution.

I looked for solutions in which $x$ and $y$ are both equal to the same power of $2$.

So suppose that $x=y=2^k$. Then the equation becomes $\bigl(2^{2k+1}\bigr)^n = \bigl(2^{2k}\bigr)^{2014}$,
$2^{(2k+1)n} = 2^{4028k}$,
$(2k+1)n = 4028k = 4*19*53k$.
Since $k$ and $2k+1$ are co-prime, it follows that $2k+1$ must be an odd divisor of $4028$, namely $19$, $53$ or $19*53 = 1007$. The corresponding values of $k$ are $9$, $26$ and $503$. That gives these three solutions to the problem:

$x=y=2^9, \ n = 4*53*9 = 1908$,

$x=y=2^{26}, \ n = 4*19*26 = 1976$,

$x=y=2^{503}, \ n = 4*503 = 2012$.

That's as far as I can go. I believe that those three solutions should be the only ones, but I can't see how to prove that there are no others.

 

[anemone]

Thanks Opalg for your participation and the partial solution, which your answer is definitely correct.

I will post the model solution below, I hope you and the readers will enjoy reading the solution.

Spoiler

Assume that $(x^2+y^2)^n=(xy)^{2014}$ holds for some positive integers $n,\,x$ and $y$. From $x^2+y^2\ge 2xy>xy$ it follows that $n<2014$. Let $d=\gcd(x,\,y)$ and $a=\dfrac{x}{d}$, $b=\dfrac{y}{d}$. Then $d^{2n}(a^2+b^2)^n=d^{2\cdot 2014}(ab)^{2014}$, which gives the equality $(a^2+b^2)^n=d^{2\cdot (2014-n)}(ab)^{2014}$. As $b$ divides the right side of this equality, $(a^2+b^2)^n$ is divisible by $b$ as well. But because $\gcd(a,\,b)=1$, also $\gcd(a^2,\,b)=1$ and $\gcd(a^2+b^2,\,b)=1$ whence $\gcd((a^2+b^2)^n,\,b)=1$, so the only possibility is $b=1$. Due to symmetry, also, $a=1$. The above equality now takes the form $2^n=d^{2\cdot(2014-n)}$. Therefore, $d=2^k$ for some $k$ and $n=2k\cdot(2014-n)$, whence $n\cdot(2k+1)=4k\cdot 1007$. Since $\gcd(2k+1,\,4)=1$ and $\gcd(2k+1,\,k)=1$, $2k+1$ divides 1007. As $1007=19\cdot 53$ and $n$ has to be positive, the possible values for $k$ are 9, 26 and 503, and the corresponding values of $n$ are 1908, 1976 and 2012.