How Many Pairs?

Assume that for some values $s, a$ we have $\left(s-1\right)\left(s\right)\left(s+1\right)\left(s+2\right) = \left(a\right)\left(a+1\right)$

Expanding the LHS yields $\left(s^2+s-2\right)\left(s^2+s\right)$ . Let $t = s^2+s-1 \Rightarrow LHS = \left(t^2-1\right)$

Hence we have $t^2 - 1 = a^2 + a$

$t^2 = a^2 + a + 1$

As $a> 0$ we have $a^2 < a^2 + a + 1 <\left(a+1\right)^2 \Rightarrow a^2 < t^2 < \left(a+1\right)^2$

This implies that there is a perfect square strictly between two consecutive perfect squares, which is clearly a contradiction and thus there can exist no pair of consecutive positive integers such that their product is the product of four consecutive integers.