# No solution for the equation $y^{n} = 2x^{n}$ for $n \geq 2$ in positive integers.

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? – Simon Markett Jun 1 '12 at 8:18
Possible duplicate of math.stackexchange.com/q/91538/6075. Although that question is only the case $n=3$, the solutions there will also apply here. – Eric Naslund Jun 1 '12 at 9:04

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. – Giuseppe Jun 1 '12 at 8:41

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." – David Loeffler Jun 1 '12 at 8:40
In fact, this was mentioned at MO as an example of using a nuke to kill a mosquito - the top example. – mixedmath Jun 1 '12 at 8:50
This has been posted a few times on MSE before ;) (See this answer ) However, I would like to point out that Fermat's Last Theorem is not strong enough for the case $n=2$. – Eric Naslund Jun 1 '12 at 9:06
@EricNaslund Yes you are right for $n=2$, we really need something stronger. – AD. Jun 1 '12 at 10:51

Hint: $2^{1/n}$ is irrational for $n \geq 2$.

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