Product of 4 consecutive integers is a perfect square,what are the possible values of their sum?

Question

I have read that product of 4 consecutive integers is less than 1from a perfect square, but this question is quite different.

Answer 1

Hint: The product of four consecutive integers plus $1$ is always a square, as you said, namely $$ n(n+1)(n+2)(n+3)+1=(n^2+3n+1)^2. $$ If the first product is already a perfect square $x^2$, then we have $x^2+1=y^2$ with $y=n^2+3n+1$. Now, how many integer solutions has $$ y^2=x^2+1. $$

Answer 2

$n(n+1)(n+2)(n+3)<n(n+1)(n+2)(n+3)+1<n(n+1)(n+2)(n+3)+2$

$n(n+1)(n+2)(n+3)+1=(n^2+3n)(n^2+3n+2)+1=((n^2+3n)^2+2(n^2+3n)+1)=(n^2+3n+1)^2$

There are only two consecutive perfect squares, they are $0$ and $1$.

The product of four consecutive integers is not a perfect square.

But if one of the consecutive integers is zero then possible solutions $\left\{-3,-2,-1,0 \right\}$,$\left\{ -2,-1,0,1\right\}$,$\left\{-1,0,1,2\right\}$,$\left\{0,1,2,3\right\}$