Are there any positive integers $n > 3$ such that $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$ is factorable in integers? If not, then why?

Question

I know that if $n = 1$ then $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$ becomes $a_1-a_1 = 0.$

If $n = 2$, then the expression becomes $a_1^2+a_2^2-2a_1a_2$, which is $(a_1-a_2)^2.$

If $n = 3$, then the expression becomes $a_1^3+a_2^3+a_3^3-3a_1a_2a_3$, which factors as $(a_1+a_2+a_3)(a_1^2+a_2^2+a_3^2-a_1a_2-a_1a_3-a_2a_3).$

If $n = 4$, I don't know how to factor it. I notice that if $a_1=a_2=a_3=a_4$, then the expression will be equal to $0$. So, I would expect something like $(a_1-a_2)^2+(a_1-a_3)^2+(a_1-a_4)^2+(a_2-a_3)^2+(a_2-a_4)^2+(a_3-a_4)^2$ to be a factor. I am not sure how to divide $a_1^4+a_2^4+a_3^4+a_4^4-4a_1a_2a_3a_4$ by $(a_1-a_2)^2+(a_1-a_3)^3+\cdots+(a_3-a_4)^2$.

Also, I see that if $a_1,a_2,a_3,a_4$ is equal to $b, b, -b, -b$ in any order, then the expression would be 0.

Similarly, for any $n$, I would expect something like $(a_1-a_2)^2 + (a_1-a_3)^2 + \cdots + (a_{n-1}-a_n)^2$ to be a factor of $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$.

Why would $n=2,3$ in $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$ be factorable but not any other $n$? Or maybe there is another $n$ such that $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$ is factorable.

I would like an answer that I can understand, without advanced vocabulary, theorems, or concepts like complex numbers beyond the basics. Please tell me if this is too much to ask for.

Does this answer your question? Find all positive integers $n$ ,such that the polynomial $ a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n $is irreducible over the real numbers - found using an Approach0 search. Note this answer there states there's a solution at the AoPS thread of ... — John Omielan, May 23 '23 at 00:44
(cont.) AM-GM Polynomial Irreducible, with post #$4$ of the search result AoPS thread Factorization also pointing to there. There are also other basic duplicate AoPS threads, e.g., Multivariable Factor Theorem, as well as related posts here, e.g., On an identity with $n$ variables, SOS: Proof of the AM-GM inequality, etc. — John Omielan, May 23 '23 at 00:50
I am afraid there is no answer without "advanced vocabulary" or "theorems." — dezdichado, May 23 '23 at 01:09
@NoChance, he’s asking if the polynomial is factorable, not the resulting integer. Also, $n$ is both the exponent and the number of variables (that way the polynomial is homogeneous), so your example isn’t what he’s talking about. — Eric, May 23 '23 at 02:51
You seem to want a polynomial factorization, not an integer factorization when $a_i$ are integers in any event, I'd bet there isn't a factorization, in general. — Thomas Andrews, May 23 '23 at 06:00

Are there any positive integers $n > 3$ such that $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2\cdots a_n$ is factorable in integers? If not, then why?

0 Answers0