What about $a_{3}=...=a_{85}=0$?

digger wrote: What about $a_{3}=...=a_{85}=0$? Are you sure?

digger wrote: What about $a_{3}=...=a_{85}=0$? $3x^{2}+2x+1=2x^{2}+(x+1)^{2}>0$ has non real roots. First note, that if $P\in\mathbb{R}[X]$ has only real roots, then so has $P'$: Let $\alpha_{1}<\cdots<\alpha_{r}$ be the real roots of $P$ with multiplicities $m_{1},\cdots,m_{r}\in\mathbb{N}$ so that $m_{1}+\cdots+m_{r}=\deg P$. Then $\alpha_{1},\cdots,\alpha_{r}$ are roots of $P'$ with multiplicities $m_{1}-1,\cdots,m_{r}-1$ (multiplicity $0$ means it's no root) and $P(\alpha_{i})=P(\alpha_{i+1})=0$ gives a root $\beta_{i}\in ]\alpha_{i},\alpha_{i+1}[$ of $P'$ for all $i=1,\cdots,r-1$. But $(m_{1}-1)+\cdots+(m_{r}-1)+(r-1)=\deg P-1=\deg P'$, so $P'$ splits in $\mathbb{R}[X]$ into linear factors and therefore has only real roots. Now if $Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{3}x^{3}+3x^{2}+2x+1$ has only real roots ($n\geq 2$), then so does $P(x)=x^{n}Q(\frac{1}{x})=x^{n}+2x^{n-1}+3x^{n-2}+\cdots \in\mathbb{R}[X]$ ($Q(0)=1\neq 0$), and so does $\frac{2}{n!}P^{(n-2)}(x)=x^{2}+\frac{4}{n}x+\frac{6}{n(n-1)}=(x+\frac{2}{n})^{2}+\frac{6}{n(n-1)}-\frac{4}{n^{2}}>0$ Contradiction!

Oh, I've mixed "any" and "some" again. I've thougt about approach with P' but never thought it could be fruitfull!

me too olorin wrote: digger wrote: What about $a_{3}=...=a_{85}=0$? $3x^{2}+2x+1=2x^{2}+(x+1)^{2}>0$ has non real roots. First note, that if $P\in\mathbb{R}[X]$ has only real roots, then so has $P'$: Let $\alpha_{1}<\cdots<\alpha_{r}$ be the real roots of $P$ with multiplicities $m_{1},\cdots,m_{r}\in\mathbb{N}$ so that $m_{1}+\cdots+m_{r}=\deg P$. Then $\alpha_{1},\cdots,\alpha_{r}$ are roots of $P'$ with multiplicities $m_{1}-1,\cdots,m_{r}-1$ (multiplicity $0$ means it's no root) and $P(\alpha_{i})=P(\alpha_{i+1})=0$ gives a root $\beta_{i}\in ]\alpha_{i},\alpha_{i+1}[$ of $P'$ for all $i=1,\cdots,r-1$. But $(m_{1}-1)+\cdots+(m_{r}-1)+(r-1)=\deg P-1=\deg P'$, so $P'$ splits in $\mathbb{R}[X]$ into linear factors and therefore has only real roots. Now if $Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{3}x^{3}+3x^{2}+2x+1$ has only real roots ($n\geq 2$), then so does $P(x)=x^{n}Q(\frac{1}{x})=x^{n}+2x^{n-1}+3x^{n-2}+\cdots \in\mathbb{R}[X]$ ($Q(0)=1\neq 0$), and so does $\frac{2}{n!}P^{(n-2)}(x)=x^{2}+\frac{4}{n}x+\frac{6}{n(n-1)}=(x+\frac{2}{n})^{2}+\frac{6}{n(n-1)}-\frac{4}{n^{2}}>0$ Contradiction!

Let $x_{1},\ldots,x_{85}$ be the roots of P. By Viete's relations $\prod_{i=1}^{85}{x_{i}}=(-1)^{85}*1=-1$ and the other two relations, so $\sum_{i=1}^{85}{x_{i}}=-2,\sum_{1\le{i}<j\le{85}}{x_{i}x_{j}}=3$ and thus $\sum_{i=1}^{85}{x_{i}^{2}}=({\sum_{i=1}^{85}{x_{i}}})^{2}-2\sum_{1\le{i}<j\le{85}}{x_{i}x_{j}}=-2<0$ so not all the roots can be real.

d.edgar wrote: Let $x_{1},\ldots,x_{85}$ be the roots of P. By Viete's relations $\prod_{i=1}^{85}{x_{i}}=(-1)^{85}*1=-1$ and the other two relations, so $\sum_{i=1}^{85}{x_{i}}=-2,\sum_{1\le{i}<j\le{85}}{x_{i}x_{j}}=3$ That's not Vieta! The correct formulas are: $1=a_{85}\prod_{i=1}^{85}(-x_{i})=-a_{85}\prod_{i=1}^{85}x_{i}$ and $2=a_{85}\sum_{j=1}^{85}\prod_{i=1,\ldots,85\atop i\not =j}(-x_{i})=a_{85}\prod_{i=1}^{85}x_{i}\sum_{j=1}^{85}{1\over x_{j}}=-\sum_{j=1}^{85}{1\over x_{j}}$ and $3=a_{85}\sum_{1\leq j<k\leq 85}\prod_{i=1,\ldots,85\atop i\not =j,k}(-x_{i})=-a_{85}\prod_{i=1}^{85}x_{i}\sum_{1\leq j<k\leq 85}{1\over x_{j}x_{k}}=\sum_{1\leq j<k\leq 85}{1\over x_{j}x_{k}}$. And now $\sum_{j=1}^{85}{1\over x_{j}^{2}}=\left(\sum_{j=1}^{85}{1\over x_{j}}\right)^{2}-2\sum_{1\leq j<k\leq 85}{1\over x_{j}x_{k}}=-2$.

olorin wrote: That's not Vieta! The correct formulas are: Uhh... At least the method was useful.