Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.

If $n = 1 + m$, where $m$ is the product of four consecutive positive
  integers, prove that $n$ is a perfect square.

Now since $m = p(p+1)(p+2)(p+3)$;

$p = 0, n = 1$ - Perfect Square
$p = 1, n = 25$ - Perfect Square
$p = 2, n = 121$ - Perfect Square

Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?

Here's another way which begins by exploiting a symmetry in the expression.
Notice that if you substitute $x=p+\frac{3}{2}$, the expression becomes
$$\left(x-\frac{3}{2}\right)\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right) + 1$$
Now see that the terms make the product of 2 differences of squares
$$\begin{align} & \quad \left(x+\frac{3}{2}\right)\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right) + 1 \\&= \left(x^2-\frac{9}{4}\right)\left(x^2-\frac{1}{4}\right) + 1 \\
&= \left(x^4 - \frac{10}{4} x^2 + \frac{9}{16}\right) + 1 \\
&= x^4 - \frac{10}{4} x^2 + \frac{25}{16} \\
&= \left(x^2 - \frac{5}{4}\right)^2 \\
&= \left(p^2 + 3p + 1\right)^2 \end{align}$$ 
which is a perfect square.
