0

Last week, my teacher asked us to prove this: Prove that $(2−1)^2−4$ is not a perfect cube for any integer I tried this approach:$y=2k$ and $y=2k+1$ but I couldn't prove it this way.

1 Answers1

3

$$(2 y - 1)^2 - 4=(2y-1+2)(2y-1-2)=(2y+1)(2y-3)$$ $(2y-3)$ and $(2y+1)$ are coprime. For euclidean algorithm

$\gcd (2y+1,2y-3)=\gcd(2y+1,4)=1$

because $(2y+1)$ is odd and is coprime with $4$

So the two numbers have no common factors. Thus or both are cubes or their product cannot be a cube.

As the difference is $4$ we see that $2y-3$ and $2y+1$ cannot be both cubes because beside the case $1^3-0^3=1$ the minimum difference between two cubes is $2^3-1^3=7$

Therefore $(2 y - 1)^2 - 4$ cannot be a cube.

Raffaele
  • 26,731