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? 

You can make the algebra simpler by starting from a better place than $m(m+1)(m+2)(m+3)$. Think symmetry!
$$\begin{align}
(m - \tfrac 3 2)(m - \tfrac 1 2)(m + \tfrac 1 2)(m + \tfrac 3 2)
&= (m^2 - \tfrac 9 4)(m^2 - \tfrac 1 4) \\
&= m^4 - \tfrac 5 2 m^2 + \tfrac{9}{16} \\
&= (m^2 - \tfrac 5 4)^2 - 1
\end{align}$$
Finally, if $m - \frac 1 2$ is an integer, $m = k + \frac 1 2$ where $k$ is an  integer.
Therefore $m^2 - \frac 5 4 = k^2 + k - 1$ is also an integer.
