If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is? If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is ?
My approach:
The given expression can be written as $(y-1)(y)(y+1)(y+k-2) + 1$ where $y = x + 2$
This gives me $(y)(y^2 - 1)(y+k-2) = n^2 - 1$ something seems similar in LHS and RHS. But not sure how to proceed next? Should I put $(y)(y+k-2) = 1$ ? although it doesn't seem right for some reason also I dont think it leads anywhere.
Any hints/explanations are appreciated!

$$(x+1)(x+2)(x+3)(x+k) + 1 = f(x)^2$$
for all $x$ then $f(x)=x^2+ax+b$ for all $x$ and some $a,b$.
Since $f(0)^2 = 6k+1$ and $f(-1)=f(-2)=f(-3)=\pm1$ so two of them must be equal and we have next $6$ options:

*

*If $f(-1)=f(-2)=1$ then $f(x) = (x+1)(x+2)+1 = x^2+3x+3$ so $f(-3) = 3$. Impossible.


*If $f(-1)=f(-2)=-1$ then $f(x) = (x+1)(x+2)-1 = x^2+3x+1$ so $f(-3)=1$ and thus $1=6k+1$ so $\boxed{k = 0}$.


*If $f(-1)=f(-3)=1$ then $f(x) = (x+1)(x+3)+1 = x^2+4x+4$ so $f(-2) = 0$. Impossible.


*If $f(-1)=f(-3)=-1$ then $f(x) = (x+1)(x+3)-1 = x^2+4x+2$ so $f(-2) = -2$. Impossible.


*If $f(-2)=f(-3)=1$ then $f(x) = (x+2)(x+3)+1 = x^2+5x+7$ so $f(-1) = 3$. Impossible.


*If $f(-2)=f(-3)=-1$ then $f(x) = (x+2)(x+3)-1 = x^2+5x+5$ so $f(-1) = 1$. Then $25=6k+1\implies \boxed{k=4}$.
