Question

Let n be a product of four consecutive positive integers then n is never a perfect square

Answer 1

Let $N=1+n$ where ${}^{\prime }{n}^{\prime }$ is a product of ${}^{\prime }{4}^{\prime }$ consecutive no.

$\Rightarrow n=a(a+1)(a+2)(a+3)$

$\Rightarrow N=1+a(a+1)(a+2)(a+3)$

$\Rightarrow N=a^4+6a^3+11a^2+6a+1$

$\Rightarrow N=(a^2+3a+1{)}^2$

${\Rightarrow }^{\prime }{N}^{\prime }$ is a perfect square

hence $n=(N-1)$ can't be a perfect square.