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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 $f(k)=l^k-b^k-2$.$f'(k)=\frac{1}{l}e^{k\log l}-\frac{1}{b}e^{k\log b}>0$ iff $e^{k(log\frac{l}{b})}>\frac{l}{b}$.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

1 Answer 1

up vote 5 down vote accepted

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.

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Thanks!So,all my effort went in vain?LOL. –  user54185 Dec 27 '12 at 18:53
    
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
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Erdos and Selfridge proved that the product of consecutive integers is never a power - see renyi.hu/~p_erdos/1975-46.pdf. –  marty cohen Dec 27 '12 at 23:53
    
Ah,thank you for linking me to the more general result. –  user54185 Dec 28 '12 at 4:26

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