# how find $a_1,a_2,…,a_n$ ,$a_i\in\mathbb R$-{0}(all possible $a_i$,i=1,2,..,n) such that $\sum_{i=1}^na_i^m=\sum_{i=1}^na_i$ for $m=1,2,…,n+1$

how find $a_1,a_2,...,a_n$ ,$a_i\in\mathbb R$-{0}(all possible $a_i$,i=1,2,..,n) such that $$\sum_{i=1}^na_i^m=\sum_{i=1}^na_i$$ $m=1,2,...,n+1$ Thanks in adnance

-
I don't understad your question. Taking $a_i=1$ for all $i$ seems to do what you want... – Mariano Suárez-Alvarez Feb 18 '13 at 18:35
@Mariano Suárez-Alvarez:i mean all possible a_i – Maisam Hedyelloo Feb 18 '13 at 18:38
Then please say so in the body of the question. – Mariano Suárez-Alvarez Feb 18 '13 at 18:39

Let $p_k=a_1^k+\dots+a_n^k$ for each $k$, and let $e_k$ be the elementary symmetric function of degree $k$ on the $a$'s.

Your hypothesis is that $$\text{p_k=p_1 for all k\in\{1,\dots,n+1\}.}\tag{\star}$$ The Newton identities tell us that $ke_k=\sum_{i=1}^k(-1)^{i-1}e_{k-i}p_i$ for all $k\in\{1,\dots,n\}$, and in our case this means that $$ke_k=p_1\sum_{i=1}^k(-1)^{i-1}e_{k-i}.$$

It follows from the hypothesis $(\star)$ that $e_k=\frac{1}{k!}p_1(p_1-1)\dots(p_1-k+1)=\binom{p_1}{i}$ for $0\leq k\leq n+1$, as one can easily check. In particular, $\binom{p_1}{n+1}=0$, and we must have $p_1\in\{0,\dots,n\}$. We also have that the $a$s are then the roots of the polynomial $$\sum_{i=0}^n(-1)^i\binom{p_1}{i}t^{n-i}.$$

Since we are looking for solutions which nn-zero components, one can easily see that this is only possible if $p=n$, and then $a_1=\dots=a_n=1$.

-
In a previous version of this answer, I had forgotten the $(n+1)$th equation. – Mariano Suárez-Alvarez Feb 18 '13 at 20:09

The expression $s_m:=\sum_{i=1}^n a_i^m$ is symmetric in the $a_i$, hence can be expressed as a polynomial in the elementary symmetric polynomials $\sigma_1,\ldots,\sigma_n$, which are the coefficients of $$\tag1X^n-\sigma_1 X^{n-1}+\sigma_2X^{n-2}\pm\ldots+(-1)^n\sigma_n=(X-a_1)\cdots(X-a_n)$$ Clearly, $s_1=\sigma_1$. Next we find $s_2=\sigma_1^2-2\sigma_2$. Since generally $\sigma_1^k=s_k+(\text{expression in other terms})+k!\sigma_k$ , one can show that the $\sigma_k$ are uniquely determined recursively if we are given a constant $s=s_1=\ldots=s_{n+1}$. Thus for given $s$ there is exactly one suitable polynomial $(1)$ and thus (up to permutation) exactly one solution $(a_1,\ldots,a_n)$. An obvious solution for $s=k$, $0\le k\le n$ is given by $a_1=\ldots=a_k=1$, $a_{k+1}=\ldots=a_n=0$. For other values of $s$, the solutions are not that simple and in fact involve complex numbers $a_i$.

Let's consider the simple case $n=2$. As seen above, $\sigma_1=s_1=s$, $\sigma_2=\frac12(\sigma_1^2-s_2)=\frac{s^2-s}2$, hence we need to solve $X^2-sX+\frac{s^2-s}2=0$. Of course the roots are $\frac {s\pm \sqrt{2s-s^2}}2$.

-
For $n=2$ and $s=\dfrac12$ you get $a_1=\dfrac{1+\sqrt{3}}{4}, \ a_2=\dfrac{1-\sqrt{3}}{4}$ right? But $a_1^3+a_2^3\neq\dfrac12$. I think in your solution you are not taking into account that $s_{n+1}=s_1$. – P.. Feb 18 '13 at 20:48
Indeed his solution is essentially equal to what I had written earlier. – Mariano Suárez-Alvarez Feb 18 '13 at 20:52