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 $25=6k+1\implies \boxed{k=4}$.
If you mean that the polynomial $f(x) = (x+1)(x+2)(x+3)(x+k)+1$ is a perfect square, i.e. $\exists g(x) \in \mathbb{Z}[x] | f(x) = g(x)^2$ then I'd try and figure out what $g(x)$ must look like. What is the degree of $f$? What then must the degree of $g$ be? Then from there you could compare the coefficients of $f$ and what $g^2$ must look like, come up with some equations, and then solve for $k$!
I.e. if $ax^4+bx^3+cx^2+dx+e = fx^4+gx^3+hx^2+ix+j$ then $a=f,b=g,c=h,d=i,e=j.$
Hint if you want to know what $k$ is:

 According to my work, $k$ should be 0 or 4.