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- #1

- Feb 14, 2012

- 3,909

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,909

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- #2

- Feb 7, 2012

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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.

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- #3

- Feb 14, 2012

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