# Product of three consecutive positive integers is never a perfect power

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!

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At lines 6, 7 there is no reason to assume $m$ and $m+1$ are perfect powers. You may be forgetting about the $2$'s. – André Nicolas Dec 27 '12 at 18:02
@AndréNicolas perhaps we can fix that? – user54185 Dec 27 '12 at 18:24
I expect one can, the nuisance is that there is more than one case to consider. I wrote out an essentially one line argument based on your initial idea. – André Nicolas Dec 27 '12 at 18:34

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.
You identified the key fact. And your proof towards the end had a distance argument. It is natural to chase down the shapes of the factors $n-1$ and $n+1$ individually. It just so happens they are better kept together. – André Nicolas Dec 27 '12 at 19:01
@user54185: By your line $5$, you had essentially the complete proof. – André Nicolas Dec 27 '12 at 19:10