2024/016) Prove that product of 4 consecutive positive integers cannot be a perfect cube.

Proof;

We  shall use the fact that if x and y are co-prime then xy is a perfect cube if both  x and y are perfect cubes.

Let the 1st number be x so the product is $x(x+1)(x+2)(x+3)$

Either x is odd or even

If x is odd then we have $GCD(x,x+2) = GCD(x+1, x+2) = GCD(x+2,x+3) = 1$

For x + 2 to be a perfect cube minimum x is 6 

So $GCD(x+2,x(+1)(x+2)) = 1$

So we need to prove that $x(x+1)(x+3)$ is not a perfect cube.

$x(x+1)(x+3) = x^3 + 4x^2 + 3$

We have $x(x+3) - (x-1)^2 = x^3+3x - x^2 -2x -1=  x-1 > 0$ as x is minimum 6 

So $x(x+1)(x+3) >= (x+1)^3$

  $x^3 + 4x^2 + 3$ is below  $(x+2)^3 = x^3 + 6x^2 + 12x + 8$

As it is between 2 cubes it cannot be a cube

Similarly we can prove taking $x +1$ when $x$ is even