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.

Answer 1

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.

Answer 2

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

Answer 3

Hint:

Apply Fermat's last theorem.

Answer 4

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.