How to factorize $n(n+1)(n+2)(n+3)+1$? How to factorize $n(n+1)(n+2)(n+3)+1$ ?
It's turned into the lowest power of $n$ ever possible but the question wants me to factorize it.
How can I do that ?

Set $\dfrac{n+n+1+n+2+n+3}4=m\iff 2n+3=2m$
$\implies n(n+1)(n+2)(n+3)+1=\dfrac{2n(2n+2)(2n+4)(2n+6)}{2^4}+1$
$=\dfrac{(2m-3)(2m-1)(2m+1)(2m+3)}{2^4}+1$
$=\dfrac{(4m^2-9)(4m^2-1)}{2^4}+1$
$=\dfrac{(4m^2)^2-2\cdot4m^2\cdot5+5^2}{2^4}=\dfrac{(4m^2-5)^2}{(2^2)^2}$
