I am trying to prove that the product of three consecutive positive integers is never a perfect power. Can anyone point to gaps in my proof and/or post an alternate solution?
Let the three positive consecutive integers be $n – 1$, $n$ and $n + 1$ and
let $(n – 1)n(n + 1) = h^k, k \ge 2$. Note that $\gcd(n – 1, n) = 1$ and $(n, n + 1) = 1$ implies that $n$ itself must be a perfect power and of the form $z^k$ (which is apparent once we look at the canonical representation of $h^k$).That means $n^2 – 1$ must be a perfect power itself and of the form $a^k$. If $n$ is odd, $n = 2m + 1$ for some $m \in \mathbb{N}$ i. e. $(n – 1)(n + 1) = 2m(2m + 2) = 2^2m(m + 1) = a^k$. Using the fact that $(m, m + 1) = 1$, $m$ and $m + 1$ must be perfect powers themselves and so $k\leq 2$. Coupled with fact that $k > 1$, we infer $k = 2$. So, $n^2 – 1 = a^2$ which implies $n$ is not a natural number.
We are left with the case when $n$ is even. Let $n = 2t + 1$. $(n + 1, n – 1) = 1$ as both are of (by the Euclidean algorithm). That means $n – 1$ and $n + 1$ are perfect powers. So, let $(n – 1) = b^k$ and $(n + 1) = l^k$. So, $l^k – b^k = 2$ for natural $l$ and $p$ and $k>1$. I will prove that the diophantine equation has no solutions.
Consider the function $f(k) = l^k – b^k – 2$. $f'(k) = \frac{1}{l}e^{k\log l} – \frac{1}{b}e^{k\log b}>0$ if and only if $e^{k(log\frac{l}{b})}>\frac{l}{b}$. Taking logarithm again, we get an equivalent condition $(k – 1)\log\frac{l}{b}>0$ which is true as $k>1$ and $l>b$ as log is a monotonically increasing function.
Is my proof correct? Please feel free to chip in with your own solutions!
Best Answer
As you observed, $n$ and $n^2-1$ are relatively prime, so perfect $k$-th powers for some $k\gt 1$. Let $n=a^k$ and $n^2-1=b^k$. Then $(a^2)^k=1+b^k$. There are not many consecutive integers that are $k$-th powers.