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 ?
 A: $$n(n+3)=n^2+3n$$
and $$(n+1)(n+2)=n^2+3n+2$$
$$\implies n(n+1)(n+2)(n+3)=(n^2+3n)[(n^2+3n)+2]+1=[(n^2+3n)+1]^2$$
A: Hint $\ $ Rearrange the product into a  $\rm\color{#0a0}{\,DS} = $ difference of squares, $ $ then add $\,\color{#c00}1$ 
$\qquad\quad  \color{#c00}n(n\!+\!1)(n\!+\!2)(\color{#c00}{n\!+\!3})\, =\, (\underbrace{\color{#c00}{n^2\!+\!3n}}_{\Large\ x^{\phantom{c}}})(\underbrace{n^2\!+\!3n\!+\!2}_{\Large\ \ x\ +\ 2a})\,\overset{\rm\color{#0a0}{DS}} =\, (\underbrace{n^2\!+\!3n\!+\!1)^2\color{#c00}{-1}}_{\Large\ \ \ (x+a)^2\,-\ a^2}$ 
A: Set $\dfrac{n+n+1+n+2+n+3}4=m\iff 2n+3=2m$
$\implies n(n+1)(n+2)(n+3)+1=\dfrac{2n(2n+2)(2n+4)(2n+6)}{2^4}+1$
$=\dfrac{(2m-3)(2m-1)(2m+1)(2m+3)}{2^4}+1$
$=\dfrac{(4m^2-9)(4m^2-1)}{2^4}+1$
$=\dfrac{(4m^2)^2-2\cdot4m^2\cdot5+5^2}{2^4}=\dfrac{(4m^2-5)^2}{(2^2)^2}$
A: Hint: Look at the first few values, for small $n$.
A: Just by running it through Wolfram, I think the simplest way to write it is $(n^2+3n+1)^2$. Many of the other ways still had a "$+1$" at the end of them.
