Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting 
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction? 

I will show that
$16m(m+1)(m+2)(m+3) = 16(m^2 + 3m + 1)^2 - 16$
It will follow that
$$m(m+1)(m+2)(m+3) = (m^2 + 3m + 1)^2 - 1$$
Proof:
\begin{align}
   16 m(m+1)(m+2)(m+3)
   &=(2m)(2m+2)(2m+4)(2m+6)\\
   & ---\text{let $2m = n-3$}---\\
   &=(n-3)(n-1)(n+1)(n+3)\\
   &=(n-3)(n+3) \cdot (n-1)(n+1) \\
   &=(n^2 - 9)(n^2 - 1) \\
   &=n^4 - 10n^2 + 9 \\
   &=(n^2 - 5)^2 - 16 \\
   & ---\text{Note $n = 2m+3$}---\\
   &=(4m^2 + 12m + 4)^2 - 16\\
   &=16(m^2 + 3m + 1)^2 - 16\\
\end{align}
addendum
I just noticed this proof.
\begin{align}
   m(m+1)(m+2)(m+3) 
   &= m(m+3) \cdot (m+1)(m+2) \\
   &= (m^2 + 3m)(m^2 + 3m + 2) \\
   &= ((m^2 + 3m + 1) - 1) \cdot ((m^2 + 3m + 1) + 1)\\
   &= (m^2+3m + 1)^2 - 1
\end{align}
