# On an identity with $n$ variables

We know the followings :

$$\color{red}{{x_1}^2+{x_2}^2-2x_1x_2}=\color{blue}{(x_1-x_2)^2}$$ $$\color{red}{{x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3}$$$$=\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_1-x_2)^2}+\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_2-x_3)^2}+\frac{1}{2}(x_1+x_2+x_3)\color{blue}{(x_3-x_1)^2}$$

I've been interested in generalizing these identities. Then, I got the followings :

$$\color{red}{{x_1}^4+{x_2}^4+{x_3}^4+{x_4}^4-4x_1x_2x_3x_4}$$$$=\frac{1}{6}\left(2({x_1}^2+x_1x_2+{x_2}^2)+(x_1+x_2)(x_3+x_4)+2x_3x_4\right)\color{blue}{(x_1-x_2)^2}$$$$+\frac{1}{6}\left(2({x_1}^2+x_1x_3+{x_3}^2)+(x_1+x_3)(x_2+x_4)+2x_2x_4\right)\color{blue}{(x_1-x_3)^2}$$$$+\frac{1}{6}\left(2({x_1}^2+x_1x_4+{x_4}^2)+(x_1+x_4)(x_2+x_3)+2x_2x_3\right)\color{blue}{(x_1-x_4)^2}$$$$+\frac{1}{6}\left(2({x_2}^2+x_2x_3+{x_3}^2)+(x_2+x_3)(x_1+x_4)+2x_1x_4\right)\color{blue}{(x_2-x_3)^2}$$$$+\frac{1}{6}\left(2({x_2}^2+x_2x_4+{x_4}^2)+(x_2+x_4)(x_1+x_3)+2x_1x_3\right)\color{blue}{(x_2-x_4)^2}$$$$+\frac{1}{6}\left(2({x_3}^2+x_3x_4+{x_4}^2)+(x_3+x_4)(x_1+x_2)+2x_1x_2\right)\color{blue}{(x_3-x_4)^2}$$

$$\color{red}{{x_1}^5+{x_2}^5+{x_3}^5+{x_4}^5+{x_5}^5-5x_1x_2x_3x_4x_5}$$$$\small=F_{1,2}\color{blue}{(x_1-x_2)^2}+F_{1,3}\color{blue}{(x_1-x_3)^2}+F_{1,4}\color{blue}{(x_1-x_4)^2}+F_{1,5}\color{blue}{(x_1-x_5)^2}+F_{2,3}\color{blue}{(x_2-x_3)^2}$$$$\small +F_{2,4}\color{blue}{(x_2-x_4)^2}+F_{2,5}\color{blue}{(x_2-x_5)^2}+F_{3,4}\color{blue}{(x_3-x_4)^2}+F_{3,5}\color{blue}{(x_3-x_5)^2}+F_{4,5}\color{blue}{(x_4-x_5)^2}$$ where $$12F_{1,2}=3({x_1}^3+{x_1}^2{x_2}+{x_1}{x_2}^2+{x_2}^3)+({x_1}^2+{x_1}{x_2}+{x_2}^2)(x_3+x_4+x_5)+(x_1+x_2)(x_3x_4+x_4x_5+x_5x_3)+3x_3x_4x_5$$ $$12F_{1,3}=3({x_1}^3+{x_1}^2{x_3}+{x_1}{x_3}^2+{x_3}^3)+({x_1}^2+{x_1}{x_3}+{x_3}^2)(x_2+x_4+x_5)+(x_1+x_3)(x_2x_4+x_4x_5+x_5x_2)+3x_2x_4x_5$$$$12F_{1,4}=3({x_1}^3+{x_1}^2{x_4}+{x_1}{x_4}^2+{x_4}^3)+({x_1}^2+{x_1}{x_4}+{x_4}^2)(x_2+x_3+x_5)+(x_1+x_4)(x_2x_3+x_3x_5+x_5x_2)+3x_2x_3x_5$$ $$12F_{1,5}=3({x_1}^3+{x_1}^2{x_5}+{x_1}{x_5}^2+{x_5}^3)+({x_1}^2+{x_1}{x_5}+{x_5}^2)(x_2+x_3+x_4)+(x_1+x_5)(x_2x_3+x_3x_4+x_4x_2)+3x_2x_3x_4$$$$12F_{2,3}=3({x_2}^3+{x_2}^2{x_3}+{x_2}{x_3}^2+{x_3}^3)+({x_2}^2+{x_2}{x_3}+{x_3}^2)(x_1+x_4+x_5)+(x_2+x_3)(x_1x_4+x_4x_5+x_5x_1)+3x_1x_4x_5$$$$12F_{2,4}=3({x_2}^3+{x_2}^2{x_4}+{x_2}{x_4}^2+{x_4}^3)+({x_2}^2+{x_2}{x_4}+{x_4}^2)(x_1+x_3+x_5)+(x_2+x_4)(x_1x_3+x_3x_5+x_5x_1)+3x_1x_3x_5$$$$12F_{2,5}=3({x_2}^3+{x_2}^2{x_5}+{x_2}{x_5}^2+{x_5}^3)+({x_2}^2+{x_2}{x_5}+{x_5}^2)(x_1+x_3+x_4)+(x_2+x_5)(x_1x_3+x_3x_4+x_4x_1)+3x_1x_3x_4$$$$12F_{3,4}=3({x_3}^3+{x_3}^2{x_4}+{x_3}{x_4}^2+{x_4}^3)+({x_3}^2+{x_3}{x_4}+{x_4}^2)(x_1+x_2+x_5)+(x_3+x_4)(x_1x_2+x_2x_5+x_5x_1)+3x_1x_2x_5$$$$12F_{3,5}=3({x_3}^3+{x_3}^2{x_5}+{x_3}{x_5}^2+{x_5}^3)+({x_3}^2+{x_3}{x_5}+{x_5}^2)(x_1+x_2+x_4)+(x_3+x_5)(x_1x_2+x_2x_4+x_4x_1)+3x_1x_2x_4$$$$12F_{4,5}=3({x_4}^3+{x_4}^2{x_5}+{x_4}{x_5}^2+{x_5}^3)+({x_4}^2+{x_4}{x_5}+{x_5}^2)(x_1+x_2+x_3)+(x_4+x_5)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3$$ Here, I have a conjecture.

Question : Is the following conjecture true?

Conjecture : For any $$n\ (\ge 2\in\mathbb N)$$ variables $$x_1,\cdots,x_n$$, the following holds.

$$\sum_{i=1}^{n}{x_i}^n-n\prod_{i=1}^{n}x_i=\sum_{1\le i\lt j\le n}(x_i-x_j)^2\sum_{k=0}^{n-2}\frac{\left(\sum_{m=0}^{k}{x_i}^m{x_j}^{k-m}\right)\cdot s_{n-2,n-2-k}(\not= x_i,x_j)}{{(n-1)\binom{n-2}{k}}}$$

where $$s_{n-2,n-2-k}(\not= x_i,x_j)=s_{n-2,n-2-k}(x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_{j-1},x_{j+1},\cdots,x_n)$$ and $$s_{n,k}(x_1,\cdots,x_n)$$ is the sum of $$\binom{n}{k}$$ products of every $$k$$ elements chosen from $$\{x_1,\cdots,x_n\}$$ with $$s_{n,0}(x_1,\cdots,x_n)=1$$.

I've already checked that this is true for $$n=2,3,4,5,6$$ and $$7$$, but I don't have any good idea to prove that. Can anyone help?

• have you tried induction on $n$? – danimal Mar 9 '15 at 11:57
• @danimal: I tried, but I could not find any good way using induction on $n$. – mathlove Mar 9 '15 at 12:09
• I have know <inequalitys> by G.Hardy,J,E,Littlewood& G.Polya, page 55，The identity of Hurwitz and Muirhead proved $$a^n_{1}+a^n_{2}+\cdots+a^n_{n}-na_{1}a_{2}\cdots a_{n}$$ can be expressed as a sum in which every term is obviously non-negative,maybe you can find their paper – math110 Mar 9 '15 at 12:37
• @math110: Thank you for your comment. So far I can find this paper whose main theorems are very similar to my conjecture. – mathlove Mar 11 '15 at 11:02

It suffices to prove, that after reducing the entire thing, there are no terms left with $x_1^a$ for any exponent $2\leq a\leq n-1$. Then it is easy to check that the coefficients of $x_1^n$ and $x_1x_2\cdots x_n$ are $1$ and $-n$ respectively. The rest follows from the symmetry of the expression.

It is easy to see why there can be no expression left containing $x_1^a x_2^b$ where $a,b\geq 2$ since this can only stem from the $(x_1-x_2)^2$-part from the subexpression $$(x_1-x_2)^2\sum_{i+j=a+b-2}x_1^ix_2^j$$ in which we can identify $$x_1^2(x_1^{a-2} x_2^b)+x_2^2(x_1^a x_2^{b-2})-2x_1x_2(x_1^{a-1}x_2^{b-1})=0$$ as the only terms resulting in $x_1^a x_2^b$ for $a,b\geq 2$. By symmetry, this shows that there will be no terms left of the form $x_i^a x_j^b$ with $a,b\geq 2$.

So it remains to show that for $2\leq a\leq n-1$ the terms of the form $$x_1^a\prod_{i\in S} x_i$$ for any given subset $S\subseteq\{2,3,...,n\}$ all cancel. For a given $a$, these can only come either from expressions of the form $$q(x_1^{a-1}+x_1^{a-2}x_2+...)(\text{products of other }x_i\text{'s})(x_1^2+x_2^2-2x_1x_2)$$ from which we get $-qx_1^a x_2(\text{products of other }x_i\text{'s})$ or from expressions of the form $$z(x_1^{a-2}+x_1^{a-3}x_2+...)(\text{products of other }x_i\text{'s})(x_1^2+x_2^2-2x_1x_2)$$ from which we get $z x_1^a(\text{products of other }x_i\text{'s})$. Now consider a specific product $$x_1^a \prod_{i\in S}x_i$$ for a fixed set $S\subseteq\{2,3,...,n\}$. This appears once in every $(x_1-x_s)^2$-part as $-qx_1^a\prod_{i\in S}x_i$ for $s\in S$. And it appears once in every $(x_1-x_t)^2$-part as $zx_1^a\prod_{i\in S}x_i$ for $t\in\{2,3,...,n\}\setminus S$.

Note that we have $|S|=n-a$ choices for $s$ and $a-1$ choices for $t$, so if these contributions are to cancel this happens iff $(a-1)z-(n-a)q=0$ so that the terms cancel if and only if $$\frac{q}{z}=\frac{a-1}{n-a}$$ This uniquely determines the ratio between the coefficients $q$ and $z$ of successive parts $q(x_1^{a-1}+x_1^{a-2}x_2+...)$ and $z(x_1^{a-2}+x_1^{a-3}x_2+...)$ for $2\leq a\leq n-1$ which actually covers all the relevant coefficients.

Finally note that, since in your general formula $q=\frac{1}{(n-1)\binom{n-2}{a-1}}$ and $z=\frac{1}{(n-1)\binom{n-2}{a-2}}$, these actually satisfy $$\frac{q}{z}=\binom{n-2}{a-2}/\binom{n-2}{a-1}=\frac{a-1}{n-a}$$ as desired.

• Thank you for your answer. I think I can get what you mean. By the way, $q=\frac{1}{(n-1)\binom{n-2}{a-1}},z=\frac{1}{(n-1)\binom{n-2}{a-2}}$, right? – mathlove Mar 11 '15 at 10:22
• @mathlove: Yes, those are $q$ and $z$. I am sorry if my rendering of it is not entirely clear. I just wrote down some principles about the terms on my paper that made sense to me, but to communicate those without giving too many or too few details to maintain a clear course, did seem quite challenging! – String Mar 11 '15 at 10:28
• BTW, I actually thought of $q,z$ as being scaled by the least common multiply of all those coefficient's denominators and moving $\frac{1}{\operatorname{(n-1)lcm}\left(\binom{n-2}{0},\binom{n-2}{1},...,\binom{n-2}{n-2}\right)}$ out in front of each part - just like you did for $n=3,4$ and $5$. After all, only the proportions are relevant in order to make the terms cancel. – String Mar 11 '15 at 10:36
• Yes, that was what I implied in my first comment. Since $q,z$ can be expressed as my first comment, $(a-1)z-(n-a)q$ does equal $0$, so the terms cancel. To be honest, it took a lot of time to understand your answer:) but I think your answer is well elaborated. Thank you! – mathlove Mar 11 '15 at 10:45
• Technical details about rather clear ideas tend to be tedious to read through. The main idea I presented here was to divide into cases that were more distinct and thus simpler to manage. You probably would have had a a simpler job simply managing those yourself than to struggle with my write-up of it ;) – String Mar 11 '15 at 10:54