The product of 3 consecutive positive integers is never a square.

Proof: if we have $a(a^2-1)=b^2$, then$\gcd(a,a^2-1)=\gcd(a^2-1,a^2)=1$, so$a=m^2, a^2-1=n^2$, where$mn=b$, and$\gcd(m,n)=1$. But $a^2-1=n^2$ has no integer solution.