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Exercise from Nathanson's book.

  • Let $n \geq 2$. Prove that the equation $y^{n}=2x^{n}$ has no solution in positive integers.

Attempt: We can write the equation as $y^{n}-x^{n} = x^{n}$. I am stuck. Did some binomial expansion and things like that but didn't work.

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    What happens if you write $y=2^ky'$ and $x=2^lx'$ with $x'$ and $y'$ odd?2012-06-01
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    Possible duplicate of http://math.stackexchange.com/q/91538/6075. Although that question is only the case $n=3$, the solutions there will also apply here.2012-06-01

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HINT: Write it instead as $\left(\frac{y}x\right)^n=2$. If this had a solution in integers, $2$ would have a rational $n$-th root. Are you familiar with a proof that $\sqrt2$ is irrational? If so, try adapting it.

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My approach would be to use uniqueness of prime factorisation on both sides of $y^n = 2x^n$, and then look at the highest power of 2 which divides both sides.

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    And then obtain a proof by reductio ad absurdum.2012-06-01
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Hint:

Apply Fermat's last theorem.

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    This suggestion reminds me of a joke Cambridge exam paper, which had a group theory question accompanied by the remark "If you use the classification of finite simple groups, you should prove it."2012-06-01
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    In fact, this was [mentioned](http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/42519#42519) at MO as an example of using a nuke to kill a mosquito - the top example.2012-06-01
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    This has been posted a few times on MSE before ;) (See [this answer](http://math.stackexchange.com/a/91757/6075) ) However, I would like to point out that Fermat's Last Theorem is not strong enough for the case $n=2$.2012-06-01
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    @EricNaslund Yes you are right for $n=2$, we really need something stronger.2012-06-01
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Hint: $2^{1/n}$ is irrational for $n \geq 2$.