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!
 A: $$(x+1)(x+2)(x+3)(x+k) + 1 = f(x)^2$$
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 $25=6k+1\implies \boxed{k=4}$.
A: So we must have for some quadratic $f$,
$$(x+1)(x+2)(x+3)(x+k) = (f+1)(f-1)$$
This implies the LHS can be split into two quadratics which differ only in the constant term, by $2$.  Clearly $f$ is monic, and $f$'s linear term can only be among $(3x, 4x, 5x)$, as combining any two of the known three terms on LHS into a quadratic gives $(x+1)(x+2), (x+1)(x+3)$ or $(x+2)(x+3)$.
Corresponding to $(3x, 4x, 5x)$, the only possibilities for $k$ are $(0,2, 4)$, by requiring the remaining quadratic factor in LHS to also have same linear term.  Now for each of these, the difference in constant term of the two quadratic factors is in order $(2, 1, 2)$, so the only solution is $k\in \{0, 4\}$.
A: 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.

