<s>[INST] Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:

Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.

I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
 [/INST] The polynomial $x^4+6x^3+11x^2+6x+1$ has symmetric coefficents - more precisely, it's called a palindromic polynomial:
$$p(x)=x^4+ax^3+bx^2+ax+1$$
The goal is to factor $p(x)$ (and show that it factors to a square of some expression). Let's start by dividing by $x^2$ and refactoring:
$$\frac{p(x)}{x^2} = q(x) = x^2+\frac{1}{x^2}+a\left(x+\frac{1}{x}\right)+b$$
It's tempting to make a substitution $y=x+\frac{1}{x}$:
$$q(x) \rightarrow q(y) = y^2+ay+(b-2)$$
This: $q(y)=0$, being a quadratic equation, is something that can be automatically solved:
$$y_{1,2}=\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}$$
and inserting the values $a=6$ and $b=11$, one obtains (note the expression under the square root is equal to zero):
$$y_1=y_2=-3$$
so that $q(y)=(y+3)^2$ - at this point one sees it's a perfect square, and due to $q(y)=\frac{p(x)}{x^2}$, basically we're done with the proof at this point.

Just to take things to their end:
Because $q(y)=0 \Leftrightarrow q(x)=0 \Leftrightarrow p(x)=0 $:
$$q(y) = (y+3)^2 = \left(x+\frac{1}{x}+3\right)^2 = \frac{(x^2+3x+1)^2}{x^2} = \frac{p(x)}{x^2}$$
hence  $x^4+6x^3+11x^2+6x+1 = (x^2+3x+1)^2$ - a perfect square indeed.
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