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 solve this without much thinking. 
By expanding the product, we find $m(m+1)(m+2)(m+3) = m^4 + 6m^3 + 11m^2 + 6m$. That's a bit bigger than $(m^2)^2 = m^4$. $(m^2+c)^2 = (m^4 + 2m^2c + c^2)$ is still too small because there is no term $m^3$. $(m^2+cm)^2 = m^4 + 2cm^3+c^2m^2$ looks better, especially if we let c = 3. 
$(m^2+3m)^2 = m^4 + 6m^3 + 9m^2$ is still just a little bit too small, by about $2m^2$. So we add another c: $(m^2+3m+c)^2 = m^4 + 6m^3 + 2cm^2 + 9m^2 + 6cm + c^2$. We let c = 1 to match the term $11m^2$ and get $(m^2+3m+1)^2 = m^4 + 6m^3 + 11m^2 + 6m + 1$. Exactly 1 more than the product, just what we wanted. So 
$m(m+1)(m+2)(m+3) = (m^2+3m+1)^2 - 1$.
That's the answer, without much thought at all.
