<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] To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.
$n^2 - 1 = (n+1)(n-1).$
OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?
Let's throw in a simple example: $1\times2\times3\times4$. Notice how $1\times4=4$ and $2\times3=6$.
What about $2\times3\times4\times5$? This time $2\times5=10$ and $3\times4=12$.
It looks like the product of the "outer" pair is $n-1$ and the product of the "inner pair" is $n+1$.
Now we know how to attack this.
Let $k$ be some integer and $m=k(k+1)(k+2)(k+3)$.
$\begin{align} m &= k(k+1)(k+2)(k+3) \\
                 &= (k+1)(k+2)\times(k(k+3)) \ \text{ (collecting inner and outer terms)}\\
                 &= (k^2 + 3k + 2) \times (k^2 + 3k) \\
                 &= ((k^2 + 3k + 1) + 1) ((k^2 + 3k + 1) - 1) \\
                 &= (k^2 + 3k + 1) ^ 2 - 1 \qquad \text{ (since }(a+b)(a-b)=a^2-b^2\text{)}.
\end{align}$
Since $k$ is an integer, $k^2 + 3k + 1$ is an integer so $m+1$ is a perfect square.
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