# 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!

• 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. Dec 27, 2012 at 18:02
• @AndréNicolas perhaps we can fix that? Dec 27, 2012 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. Dec 27, 2012 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. Dec 27, 2012 at 19:01
• @user54185: By your line $5$, you had essentially the complete proof. Dec 27, 2012 at 19:10