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

Question

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.

What happens if you write $y=2^ky'$ and $x=2^lx'$ with $x'$ and $y'$ odd? — Simon Markett, Jun 01 '12 at 08:18
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. — Eric Naslund, Jun 01 '12 at 09:04

score 10 · Accepted Answer · answered Jun 01 '12 at 08:18

10

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.

answered Jun 01 '12 at 08:18

Brian M. Scott

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score 7 · Answer 2 · answered Jun 01 '12 at 08:35

7

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.

answered Jun 01 '12 at 08:35

Old John

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And then obtain a proof by reductio ad absurdum. – agt Jun 01 '12 at 08:41

score 3 · Answer 3 · answered Jun 01 '12 at 08:25

3

Hint:

Apply Fermat's last theorem.

answered Jun 01 '12 at 08:25

AD - Stop Putin -

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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 01 '12 at 08:40
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In fact, this was mentioned at MO as an example of using a nuke to kill a mosquito - the top example. – davidlowryduda Jun 01 '12 at 08:50
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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 01 '12 at 09:06
@EricNaslund Yes you are right for $n=2$, we really need something stronger. – AD - Stop Putin - Jun 01 '12 at 10:51

score 2 · Answer 4 · answered Jun 01 '12 at 08:20

2

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

answered Jun 01 '12 at 08:20

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

4 Answers4