<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] Suppose that you have 4 consecutive numbers $a, b, c, d$. They can be expressed as $a=t-\frac{3}{2}$, $b=t-\frac{1}{2}$, $c=t+\frac{1}{2}$ and $d=t+\frac{3}{2}$ for some number $t$.
Now,
$$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = t^2 - \left(\frac{3}{2}\right)^2 = t^2 - \frac{9}{4}$$
and
$$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = t^2 - \left(\frac{1}{2}\right)^2 = t^2 - \frac{1}{4}$$
If we define $y = t^2 - \frac{5}{4}$, we have 
$$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = y - 1$$
and
$$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = y + 1$$
Furthermore, since the LHS in both cases is an integer, it is clear that $y$ is an integer.
So the product of all four numbers is 
$$ abcd = (y - 1)(y+1) = y^2 - 1 $$
one less than the square of an integer.
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