# $n(n+1)(n+2)$ is not a perfect power [duplicate]

We have $n,n+1, n+2 \in \mathbb Z^+$ Their product can't be a whole exponentiation. Why?

I noticed that $gcd(n,n+1)=1$ and $gcd(n+1,n+2)=1$ This could be a good starting point in the proof. But how do I proceed? Thanks

• Do you mean that $n(n+1)(n+2)$ can't be a perfect power of an integer, in the form $a^m$ for some $m > 1$? Mar 4, 2015 at 22:17
• That's what I meant, thanks and sorry about that, I'm spanish so my english (especially my mathematical english) isn't the best Mar 4, 2015 at 22:24
• @user176791 Completely understood! I just wanted to make sure everyone was on the same page. Mar 4, 2015 at 22:59

Because you don't check the links I gave you, I write this answer:

First of all $n+1$ and $n(n+2)=(n+1)^2-1$ are relatively prime, so they are both perfect $k$-th powers for some $k>1$. Let $n+1=a^k$ and $(n+1)^2−1=b^k$. Then $(a^2)^k=1+b^k$. hence $$(a^2)^k-b^k=1$$ but the are not many positive consecutive powers.

Proof by @Andrés Nikolas

In 1975, Erdos and Selfridge proved that the product of consecutive integers is never a power.

$n(n + 1)(n + 2) = n^3 + 3n^2 + 2n$. Obviously this is divisible by $n$. So let's divide it by $n$: we get $n^2 + 3n + 2$. If $n > 2$, then $n^3 - n^2 > 3n + 2$, meaning that $n^2 + 3n + 2$ is between $n^2$ and $n^3$. Therefore $n^2 + 3n + 2$ can't be a perfect power of $n$, and neither can $n^3 + 3n^2 + 2n$.
This still leaves you $n = 1$ and $n = 2$ to check, but that's done easily enough.
• This post doesn't answer the question : $n(n+1)(n+2)$ can't be a perfect power of an integer, in the form $a^m$ , because there is perfect powers between $n^2$ and $n^3$ and it's not very clear your arguments, so the best answer is here Mar 4, 2015 at 22:39
• This is not exactly right: in order that $n\cdot(n+1)(n+2)$, $(n+1)(n+2)$ has not to be a power of $n$. Mar 4, 2015 at 22:39