http://mathforum.org/library/drmath/view/68049.html

What I wrote in the exam : Define $r_1, r_2 , r_3 , r_4$ to be roots of $ x^{4}+ax^{3}+bx^{2}+cx+d $ , then by Vieta's relations $P(a,b,c,d) \\ = P(-\sum_{i=1}^{4}r_i \;\: ,\; \sum_{1 \leq i<j \leq 4}^{ }r_i r_j \;\: ,\; -\sum_{1 \leq i < j < k \leq 4}^{ } r_i r_j r_k \;\: ,\; r_1 r_2 r_3 r_4) \\ = Q(r_1 , r_2 , r_3 ,r_4)$ Then condition is $Q(x,y,z,t) \geq 0$ if $x,y,z,t \in \mathbb R$ and it becomes negative when $z,t$ are two conjugate non-real complex numbers (here we should note that $a,b,c,d \in \mathbb R$) . Let $x,y,r \in \mathbb R^$ and $z=\zeta \: ,\: t= \bar{\zeta }$ where $|\zeta |=r^2$ ,Then $Q(x,y,\zeta , \bar{\zeta} ) $ becomes a polynomial because $\zeta = r . \theta$ and $\bar{\zeta}=r . \frac{1}{\theta}$ for some $\theta$ on unit circle . Let $\zeta$ varies on complex numbers with norm $r^2$ .For every such $\zeta $ , $Q(x,y,\zeta , \bar{\zeta} ) <0$ ,except when $ \zeta \in \mathbb R$ , i.e $\zeta = \pm r$ . In this two exceptions we should have $Q(x,y,\pm r,\pm r) \geq 0$ , But according to continuity of $Q(x,y,\zeta , \bar{\zeta} ) $ , $Q(x,y,\pm r,\pm r)$ can not be positive because if it become positive then there is a $\epsilon$-neighbor of $r$ that it remains positive on it , which is absurd . Hence $H(r)=Q(x,y,r,r)=0$ for all real $r$ . $H(r)$ vanish at every $r$ and all $x,y$ $\implies$ $Q(i,-i,1,1)=0$ where $i^2 =-1$ , contradiction !

My solution (What I wrote in the exam... with more details (3 pages!)) If $Q(x)=(x^2+mx+n)(x^2+px+q)$ then $P(a,b,c,d)=P(m+p,mp+n+q,np+mq,nq)=R(m,n,p,q)$ Then condition is equivalent to $R(m,n,p,q)\geq0\Longleftrightarrow m^2-4n\geq0, \ p^2-4q\geq0$ Let $p^2-4q>0$ so $R(m,n,p,q)\geq0\Leftrightarrow m^2-4n\geq0$ thus $m^2-4n|R(m,n,p,q)$. Let $k$ be the greatest positive integer s.t. $(m^2-4n)^k|R(m,n,p,q)$. Let $p^2-4q<0$ so $R(m,n,p,q)<0 \ (\forall m,n\in \mathbb{R})$. Thus $k$ is even. Let $p^2-4q>0$ so $R(m,n,p,q)\geq0\Leftrightarrow m^2-4n\geq0$. Thus $k$ must be odd, contradiction! Is it true?

hadikh wrote: Let $p^2-4q>0$ so $R(m,n,p,q)\geq0\Leftrightarrow m^2-4n\geq0$ thus $m^2-4n|R(m,n,p,q)$. Would you please prove this part? Thanks.

goodar2006 wrote: hadikh wrote: Let $p^2-4q>0$ so $R(m,n,p,q)\geq0\Leftrightarrow m^2-4n\geq0$ thus $m^2-4n|R(m,n,p,q)$. Would you please prove this part? Thanks. Fix $p,q$ s.t. $p^2-4q>0$ and let $m\in \mathbb{R}$ be arbitrary constant. Let $R(m,n,p,q)=T(n)$ If $n<\frac{m^2}{4}$ then $m^2-4c>0$ so $T(n)\geq0$. If $n>\frac{m^2}{4}$ then $m^2-4c<0$ so $T(n)<0$. Thus $n=\frac{m^2}{4}$ is a root of $T$ and $m^2-4n$ is a factor of $T$. $m$ was arbitrary constant, so $m^2-4n$ is a factor of $R$.

Could you explain your solution, goodar2006 ?

goodar2006’s solution was wrong, it is possible to have $Q(b,\frac{b^2}{4})=0$ for all non-negative $b$. He reduced the equation to a quadratic equation, then $Q$ is just $b^2-4d$, we cannot prove by contradiction. We still don’t know if $P$ exists.

Please, do provide a polynomial $Q(b, d)$ where $Q(b, \frac{b ^ 2}{4}) = 0$ only for nonpositive real numbers $b$. I think you haven't completely understood the problem or my solution yet. Please read it more carefully.

Thanks for replying, I understand your answer now. Let $f(b)=Q(b,\frac{b^2}{4})=0$, there is no polynomial $f$ that $f(b)=0$ only for $b\le0$. Because it's also valid for $b>0$, we cannot use $Q$ to distinguish whether the equation has four real roots. It's indeed a really neat answer.