An arbitrary product of four consecutive integers can be expressed as $(n-1)n(n+1)(n+2)$. The conditions of the problem dictate that $n>1$. This product may be rewritten as $(n-1)(n+2)n(n+1) = (n^2+n-2)(n^2+n)$. This means that the product of four consecutive integers has the form $(m-2)m$, i.e. as the product of two positive integers that differ by two. We show that no perfect square has this form. If the sequence of integers form a perfect square $x^2 = x x$, it is clear that x cannot be a prime. This means that x must be divisible by some number $a>1$ resulting in a factor $x/a$ where $(x/a)(x/a+2) = x^2 \Rightarrow x/a+2 = ax$. From this last expression we have.

\[ \begin{eqnarray*} ax-\frac{x}{a} &=& 2, \\ \frac{a^2x-x}{a} &=& 2, \\ x &=& \frac{2a}{a^2-1}. \end{eqnarray*}\]

There is no value of a that provides an integer solution to the above equation since for $a = 2$, $x=4/3$ and for $a > 2$ the numerator is less than the denominator.