Product of Three Consecutive Integers is never Perfect Power

Theorem

Then:

$\paren {n - 1} n \paren {n 1}$

cannot be expressed in the form $a^k$ for $a, k \in \Z$ where $k \ge 2$.

That is, the product of $3$ consecutive (strictly) positive integers can never be a perfect power.

Aiming for a contradiction, suppose $\paren {n - 1} n \paren {n 1} = a^k$ for $a, k \in \Z$ where $k \ge 2$.

We have that:

$\gcd \set {n − 1, n} = 1 = \gcd \set {n, n 1}$

Thus $n$ must itself be a perfect power of the form $z^k$ for some $z \in \Z$.

That means $\paren {n - 1} \paren {n 1} = n^2 - 1$ must also be a perfect power of the same form.

Let:

$n = r^k$ and $n^2 − 1 = s^k$

for $r, s \in \Z$.

Then:

$\paren {r^2}^k = 1 s^k$

But the only consecutive integers that are $k$th powers are (trivially) $0$ and $1$.

Hence by Proof by Contradiction there can be no such $n$.

$\blacksquare$