Integer solutions to $(a_1+a_2+\cdots+a_n)^n=a_1a_2\cdots a_n$?

Are there any positive nonzero integer solutions to $(a_1+a_2+\cdots+a_n)^n=a_1a_2\cdots a_n$ for $n>1$?

If it helps, there are no solutions for $n=2$ because, otherwise, if $a_1$ and $a_2$ were both odd then $(a_1+a_2)^{2}$ would be even and $a_1a_2$ would be odd, and if $a_1$ were odd and $a_2$ were even or vice versa then $(a_1+a_2)^{2}$ would be odd and $a_1a_2$ would be even, and if $a_1$ and $a_2$ were both even then $a_1/2$ and $a_2/2$ would also solve the equation so there would exist a solution with at least one of them odd.

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$(a_1+...+a_n)^n \geq (a_1)(a_2)\cdots(a_n)=a_1\cdot \cdots \cdot a_n$ The equality only holds iff $a_1 = \cdots = a_n = 0$ –  Jineon Baek Sep 11 '11 at 7:08
It is not clear what you mean. When $n=3$, is the left hand side supposed to be $(a_1+a_2+a_3)^3$ or is it supposed to be $(a_1+a_2a_3)^3$? It doesn't really matter this time (you won't have any solutions in either case), but it is a bit confusing. Also, please use \cdot to get a centered dot for multiplication. A period is easy to mix with something else. –  Jyrki Lahtonen Sep 11 '11 at 7:22
But the impossibility of this equation is easy. Let $S$ be the largest of the integers $a_i,i=1,2,\ldots,n$. Then the right hand side is $\le S^n$, and the left hand side is $\ge (S+1)^n$. Irrespective of which of the alternative meanings you want the l.h.s. to have :-) –  Jyrki Lahtonen Sep 11 '11 at 7:24
Indeed, for positive $a_i$ we have $$(a_1+\cdots+a_n)^n>n! a_1\cdots a_n$$ by the multinomial theorem. –  anon Sep 11 '11 at 7:25

The answer follows immediately from AM-GM inequality. You are given that $a_1,a_2,\ldots,a_n$ are positive. From AM-GM, it follows that $$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1a_2\cdots a_n}.$$ i.e. $$\left( a_1 + a_2 + \cdots + a_n \right)^n \geq n^n \left(a_1a_2\cdots a_n \right).$$ Given that $\left( a_1 + a_2 + \cdots + a_n \right)^n = a_1 a_2 \cdots a_n$, we get $a_1 a_2 \cdots a_n \geq n^n a_1 a_2 \cdots a_n$. Hence, we have $n^n \leq 1$. Hence, it is true only when $n=1$.

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Ah man, your $n^n$ beats my $n!$. :( –  anon Sep 11 '11 at 10:15

Someone might as well write an answer. This one is based on the multinomial theorem I linked above but explained casually. Whenever a sum of terms is raised to a power (say $n$), one can foil it out to a sum of products of the original terms, e.g. $(x+y)^2=x\cdot x+2x\cdot y+y\cdot y$. Each product can be written with exactly $n$ terms, each chosen from a distinct copy of the sum. Order matters: since you can choose one $x$ and one $y$ from $\{x,y\}$ in two different orders, the product $xy$ appears in the expansion of $(x+y)^2$ twice.

Generalizing, the sum $(a_1+a_2+\cdots+a_n)^n$, when foiled, will contain a product of the form $a_1a_2\cdots a_n$. In fact, it will contain $n!$ copies of the product because each order of picking $a_1,a_2,\dots,a_n$ from $\{a_1,a_2,\dots,a_n\}$ is a permutation, and there are $n!$ permutations of $n$ distinct objects. Hence we have $(\sum)^n=n!a_1a_2\cdots a_n+S$, where $S$ is the sum of all the other products that come out of the foiling. Since each of the $a_i$ are positive by hypothesis, each product in the sum $S$ is positive, hence $S>0$, which entails the strict inequality $$(a_1+a_2+\cdots+a_n)^n>n!\cdot a_1a_2\cdots a_n.$$ This prevents equality from ever occuring, whole numbers or otherwise.

P.S. If the question is actually about $(a_1+a_2a_3\cdots a_n)$, which I highly doubt, then we can just make the simple enough substition $a=a_1$, $b=a_2a_3\cdots a_n$ and we're back at the $n=2$ case.

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