<s>[INST] Proving on the equation $(x^2+mx+n)(x^2+px+q)=0$ Find all real numbers k such that if $a,b,c,d \in \mathbb R$ and $a>b>c>d \geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that the following equation has 4 distinct real solutions :
$(x^2+mx+n)(x^2+px+q)=0$
Here all i did :
$(x^2+mx+n)(x^2+px+q)=0$
$\Leftrightarrow x^2+mx+n=0$ or  $x^2+px+q=0$
so I think the four real solutions, if any, of the equation can only be :

*

*$ x= \frac{\sqrt{m^2-4n} -m}{2} $


*$ x= \frac{-\sqrt{m^2-4n} -m}{2} $


*$ x= \frac{\sqrt{p^2-4q} -p}{2} $


*$ x= \frac{-\sqrt{p^2-4q} -p}{2} $
So I think to solve the problem we just need to find all real numbers k such that if $a,b,c,d \in \mathbb R$ and $a>b>c>d \geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that
$p^2 \geq 4q$ and $ m^2 \geq 4n$
We can prove that for $k \geq 4 $ it is absolutely true. So are there any other satisfying $k $ values ? I'm not entirely sure. Hope to get help from everyone. Thanks very much !
 [/INST] The question needs to be answered in three steps: First, we need to place constraints on $m,n,p,q$ so that all four roots are real.  Second, we need to add constraints so that all four roots are distinct.  Lastly, we need to find $k$ such that for all $a>b>c>d>k$, at least one permutation exists where all the constraints hold true.  Note that by symmetry, permuting $m$ with $p$ and $n$ with $q$ will not change the constraints.
As stated in the post, in order for all four roots to be real, we require $m^2\geq 4n$ and $p^2\geq 4q$.  If equality holds in either constraint, we won't have distinct roots, so our constraints are $m^2>4n$ and $p^2>4q$.  Suppose $k=4-\delta$ for some $\delta>0$. Then, If all four of $m,n,p,q\in(4-\delta,4)$ these constraints cannot hold.  As a result, we require at the very least $k\geq 4$, along with possibly further constraints.
The above constraint guarantees that the two roots defined by $m,n$ are different from each other, and that the two roots defined by $p,q$ are different from each other.  We still need to exclude the remaining pairs of roots.  Let $y,z\in\{-1,1\}$ so that the remaining pair of roots equations can be set up as $\frac{-y\sqrt{m^2-4n}-m}{2}=\frac{-z\sqrt{p^2-4q}-p}{2}$. We will then develop this:
$$-y\sqrt{m^2-4n}-m=-z\sqrt{p^2-4q}-p$$
$$y\sqrt{m^2-4n}+m=z\sqrt{p^2-4q}+p$$
$$y\sqrt{m^2-4n}-z\sqrt{p^2-4q}=p-m$$
$$m^2-4n+p^2-4q-2yz\sqrt{(m^2-4n)(p^2-4q)}=m^2-mp+p^2$$
$$-2yz\sqrt{(m^2-4n)(p^2-4q)}=4n+4q-mp$$
$$4m^2p^2-16m^2q-16np^2+64nq=16n^2+16q^2+m^2p^2+32nq-8mnp-8mpq$$
$$3m^2p^2-16m^2q-16np^2+32nq-16n^2-16q^2+8mnp+8mpq=0$$
$$-16q^2+(-16m^2+32n+8mp)q+(3m^2p^2-16np^2-16n^2+8mnp)=0$$
Let $\beta=\sqrt{m^2-4n}$.  Our quadratic is now (after some algebra): $2q^2+(2\beta^2+4n-mp)q+(-\frac{3}{8}\beta^2p^2+\frac{1}{2}np^2+2n^2-mnp)=0$.  The discriminant is $4\beta^4+16n^2+m^2p^2+16\beta^2n-4\beta^2mp-8mnp+3\beta^2p^2-4np^2-16n^2+8mnp=4\beta^4+16\beta^2n-4\beta^2mp+3\beta^2p^2+m^2p^2-4np^2=4\beta^2(4n-mp)$.  This has no real roots if $mp>4n$, which means that the constraints $mp>4n, mp>4q, m^2>4n, p^2>4q$ are sufficient to guarantee that the roots are real and distinct.  Now, for $k\geq4$, note that the permutations of constraints $4\leq q,n<m,p$ are sufficient to guarantee that each of the four constraints holds true.  As such, 4 is the minimum possible value of k.
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