Given $$ P (x) = ax ^ 4 + bx ^ 3 + cx ^ 2 + dx + e = a (x ^ 2 + p_1x + q_1) (x ^ 2 + p_2x + q_2) $$ for some $ a, b, c, d, e, p_1, q_1, p_2, q_2 \in \mathbb R$, prove that $ P (x) $ can be reduced to the form $$ \frac {(M_1 + N_1t ^ 2) (M_2 + N_2t ^ 2)} {(\gamma t + 1) ^ 4} $$ by replacing $$ x = x (t) = \frac {\alpha t + \beta} {\gamma t + 1} $$

I could prove it only for the case when $ P (x) $ does not have all real roots. If there is at least one complex root $ z $, that is, its pair is the conjugate root $ \bar {z} $ and then if $$ x ^ 2 + px + q = (x-z) (x-z_1) $$ where $ p $ and $ q $ are real numbers, then $ z_1 = \bar {z} $. This implies that $ p_1, q_1, p_2, q_2 $ are numbers uniquely determined by $ a, ..., e $. This greatly simplifies the situation, so by extracting the inequality $ p_1 ^ 2-4q_1 <0 $, one can prove the required. If we have 4 real roots $ x_1, x_2, x_3, x_4 $, then in the equality $ (x-x_i) (x-x_j) = M_1 + N_1t ^ 2 $ the numbers $ i $ and $ j $ can be any of the set $ \left \lbrace1,2,3,4 \right \rbrace $. How to be in this case, I do not know.

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We homogenize the polynomial in question, and obtain quadratic forms $x^2 + p_1 xy + p_2 y^2, x^2 + q_1 xy + q_2 y^2$. We are done if we can find an element in $\text{GL}_2(\mathbb{R})$ which diagonalizes both of them.

We can certainly transform the first quadratic form to $x^2 \pm y^2$ via an element of $\text{GL}_2(\mathbb{R})$. The first case occurs precisely when the quartic has a non-real root, and the question is reduced to finding an element in the standard orthogonal group

$$\displaystyle O_2(\mathbb{R}) = \left\{\begin{pmatrix} \cos(t) & \sin(t) \\ \pm \sin(t) & \mp \cos(t) \end{pmatrix} : t \in [0, 2\pi) \right\},$$

say $T(t)$, which diagonalizes $x^2 + q_1 xy + q_2 y^2$. If $T(t) = \left(\begin{smallmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{smallmatrix}\right)$ say, then we have

$$\displaystyle (\cos(t) x + \sin(t) y)^2 + q_1 (\cos(t) x + \sin(t) y)(-\sin(t)x + \cos(t)y) + q_2 (-\sin(t)x + \cos(t)y)^2$$ whose $xy$-coefficient is given by

$$\displaystyle \sin(2t) + q_1 \cos(2t) - q_2 \sin(2t).$$

We can always find a zero for this function, hence it is always possible to diagonalize $x^2 + q_1 xy + q_2 y^2$.

Next we assume that $F(x,y) = y^4f(x/y)$ is totally real. It is standard that there exists an $A \in \mathbb{R}$ such that $F$ is $\text{GL}_2(\mathbb{R})$-equivalent to $F_A(x,y) = x^4 + Ax^2 y^2 + y^4$. We can factor $F_A(x,y) = (x^2 - uy^2)(x^2 - vy^2$ for $u,v > 0$ over $\mathbb{R}$, as desired.

  • $\begingroup$ Unfortunately, I am not yet so erudite in mathematics in order to work with groups and I need a solution within the framework of the differential calculus and the beginning of the integral $\endgroup$ – Paul Ivanov May 13 at 18:31
  • $\begingroup$ I think this is a nice idea. However, I have some questions concerning the last paragraph of your answer: What are $F$ and $f$? What is "totally real"? Where can a proof of that "standard" result be found? Also, you cannot factor $F_A(x,y) = x^4 + Ax^2 y^2 + y^4$ the way you did if $|A|<2$. $\endgroup$ – Iosif Pinelis May 14 at 13:57
  • $\begingroup$ @IosifPinelis "totally real" means that $f$ has four real roots. If $f$ has four real roots then its discriminant is positive, so it can be put into the shape I described. As you say sometimes $x^4 + Ax^2 y^2 + y^4$ is totally imaginary: i.e., has no real roots at all (and of course sometimes it ramifies). $\endgroup$ – Stanley Yao Xiao May 14 at 14:19
  • $\begingroup$ Thank you for answering one or two of my four questions/concerns. Still, what are $F$ and $f$, and where can a proof of that "standard" result be found? $\endgroup$ – Iosif Pinelis May 14 at 18:46
  • $\begingroup$ Also, as seen from my answer, there are exceptions when the desired representation is not possible. Why don't they appear in your answer? $\endgroup$ – Iosif Pinelis May 14 at 18:48

$\newcommand{\ga}{\gamma} \newcommand{\be}{\beta} \newcommand{\al}{\alpha} $ Without loss of generality, $a=1$. You already know how to deal with the case when $P(x)$ has a non-real root, and this case was also proved by Stanley Yao Xiao.

Consider now the situation when all the four roots of $P(x)$ are real. The trivial solution corresponds to the case when the transformation $$t\mapsto x=g(t):=\frac {\al t+\be}{\ga t+1}$$ is constant (that is, when $\al=\ga=0$); apparently, you don't need this trivial solution, and so, it will be excluded from the further consideration.

By shifting, without loss of generality (wlog) the roots of $P(x)$ are $a,-a,b,c$ for some real $a,b,c$, so that \begin{equation*} P(x)=(x^2-a^2)(x-b)(x-c); \tag{0} \end{equation*} details on this have been added at the end of this answer. Under the transformation $x\leftrightarrow g(t)$, the multiset $\{a,-a,b,c\}$ of the roots of $P(x)$ must be in a one-to-one correspondence with the multiset $\{u,-u,v,-v\}$ of the roots $t$ of the polynomial $Q(t):=(M_1+N_1t^2)(M_2+N_2t^2)$ -- this is a necessary and sufficient condition for your desired representation. Indeed, if we have such a match, that is, such a one-to-one correspondence of the multisets of the roots, then \begin{equation*} P(x)=R(t):=\frac{(M_1+N_1t^2)(M_2+N_2t^2)}{(\gamma t+1)^4} \tag{1} \end{equation*} for $x=g(t)$, where
\begin{equation*} N_1=P(\al),\quad N_2=1,\quad M_1=-u^2 N_1,\quad M_2=-v^2 N_2. \tag{2} \end{equation*}

Let us now get to the matching business between the multisets $\{a,-a,b,c\}$ and $\{u,-u,v,-v\}$, so that \begin{equation*} \{g(u),g(-u),g(v),g(-v)\}=\{a,-a,b,c\}. \tag{3} \end{equation*} Note that one always has at least one of the following two cases:

Case 1: $D_1:=(a^2 - b^2) (a^2 - c^2)\ge0$ or

Case 2: $D_2:=a (a + b) (a - c) (b - c)>0$.

Consider first Case 1. Then we have (3) (in fact, more specifically, here we have $(g(u),g(-u),g(v),g(-v))=(a,-a,b,c)$) with \begin{align*} \al&=\frac{\sqrt{D_1}+a^2+bc}{b+c}, \\ \be&=\frac{-\sqrt{D_1}+a^2+bc}{b+c}, \\ \ga&=1,\\ u&=\frac{-\sqrt{D_1}+a^2+b c}{a(b+c)}, \\ v&=-\frac{2\sqrt{D_1}-2 a^2+b^2+c^2}{(b-c) (b+c)}, \end{align*} and hence, with $M_1,N_1,M_2,N_2$ as in (2), we have (1) -- except when one of the denominators in the latter display equals $0$.

Consider now those exceptional subcases of Case 1:

Subcase 1.0: $b+c=0$. Then $P(x)=(x^2-a^2)(x^2-b^2)$, so that (1) holds with $x=t$, that is, with $\al=\ga=0$ and $\be=1$.

Subcase 1.1: $b=c$.

Subsubcase 1.1.0: $b=c\ne0\ \&\ a\ne0$. Then $\al=a^2/b, \be=b, \ga=1,u=b/a, v =0$ will do.

Subsubcase 1.1.1: $b=c=0\ \&\ a\ne0$. Then $\al=a, \be=0, \ga=0,u=1, v =0$ will do.

Subsubcase 1.1.2: $b=c\ne0\ \&\ a=0$. Then $P(x)=x^2(x-b)^2$, and here -- an exception among the exceptions! -- (1) cannot hold for any $M_1,N_1,M_2,N_2$.

Subsubcase 1.1.3: $b=c=0\ \&\ a=0$. This is covered by Subcase 1.0.

Subcase 1.2: $b\ne c$. Then $\al=\frac{2 b c}{c-b}, \be=0, \ga=\frac{b+c}{c-b},u=0, v =1$ will do.

In Case 2, we have (3) (in fact, more specifically, here we have $(g(u),g(v),g(-u),g(-v))=(a,-a,b,c)$) with \begin{align*} \al&=\frac{a (b+c)+\sqrt{2} \sqrt{D_2}}{2 a+b-c}, \\ \be&=\frac{a (b+c)-\sqrt{2} \sqrt{D_2}}{2 a+b-c}, \\ \ga&=1,\\ u&=-\frac{2 a^2+a (b-3 c)+2 \sqrt{2} \sqrt{D_2}+b^2-b c} {(a-b) (2a+b-c)} \\ v&=\frac{2 a^2+a (b-3 c)+2 \sqrt{2} \sqrt{D_2}+b^2-b c}{(a-b)^2 (a+c) (2 a+b-c)} \\ &\times\left(-a^2+2 \sqrt{2} \sqrt{D_2}-3 a b+3 a c+b c\right), \end{align*} and hence, with $M_1,N_1,M_2,N_2$ as in (2), we have (1) -- -- except when one of the denominators in the latter display equals $0$.

Consider now those exceptional subcases of Case 2:

Subcase 2.1: $c=2a+b$.

Subsubcase 2.1.0: $c=2a+b\ \&\ b\ne-3a$. Then $\al=\frac{1}{2} (3 a+b)$, $\be=\frac{a+b}{2}$, $\ga=0$, $u=\frac{a-b}{3 a+b}$, $v=-1$ will do.

Subsubcase 2.1.1: $c=2a+b\ \&\ b=-3a$. Then $\al=2a$, $\be=-a$, $\ga=0$, $u=1$, $v=0$ will do.

Subcase 2.2: $b=a\ \&\ c\ne2a+b$.

Subsubcase 2.2.0: $b=a\ \&\ c\ne2a+b=3a\ \&\ c\ne-a$. Then $\al=\frac{3 a c-a^2}{3 a-c}$, $\be=a$, $\ga=1$, $u=0$, $v=\frac{c-3 a}{a+c}$ will do.

Subsubcase 2.2.1: $b=a\ \&\ c\ne2a+b=3a\ \&\ c=-a$. This is covered by Subcase 1.0.

Subcase 2.3: $c=-a\ \&\ b\ne a\ \&\ c\ne2a+b$, so that $b\ne-3a$. Then $\al=\frac{a (a+3 b)}{3 a+b}$, $\be=-a$, $\ga=1$, $u=\frac{3 a+b}{b-a}$, $v=0$ will do.

Thus, your desired representation is always possible (and in fact is given here explicitly), except when $P(x)$ has two distinct real roots, each of multiplicity $2$ -- as in Subsubcase 1.1.2.

Added details on (0): It has just been shown that for any real $a,b,c$ there exist real $\al,\be,\ga,M_1,N_1,M_2,N_2$ such that for all real $t$ (with $\ga t+1\ne0$) we have \begin{equation*} Q(w):=(w^2-a^2)(w-b)(w-c)=\frac{(M_1+N_1t^2)(M_2+N_2t^2)}{(\ga t+1)^4} \end{equation*} with \begin{equation*} w=\frac {\al t+\be}{\ga t+1}. \end{equation*} Now we can write
\begin{equation*} P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)=(w^2-a^2)(w-b)(w-c) =\frac{(M_1+N_1t^2)(M_2+N_2t^2)}{(\ga t+1)^4} \end{equation*} for real $x_1,x_2,x_3,x_4$, where $w:=x-d$, $d:=\frac12\,(x_1+x_2)$, $a^2:=d^2=x_1x_2\ge0$, $b:=x_3-d$, $c:=x_4-d$. At the same time, we have \begin{equation*} x=w+d=\frac {\al_1 t+\be_1}{\ga t+1} \end{equation*} with $\al_1:=\al+\ga d$ and $\be_1:=\be+d$. So, for any real $x_1,x_2,x_3,x_4$ there exist real $\al_1,\be_1,\ga,M_1,N_1,M_2,N_2$ such that for all real $t$ (with $\ga t+1\ne0$) we have \begin{equation*} (x-x_1)(x-x_2)(x-x_3)(x-x_4) =\frac{(M_1+N_1t^2)(M_2+N_2t^2)}{(\ga t+1)^4} \end{equation*} with \begin{equation*} x=\frac {\al_1 t+\be_1}{\ga t+1}, \end{equation*} as desired.

  • $\begingroup$ @PaulIvanov : I have now given a completely different solution. $\endgroup$ – Iosif Pinelis May 14 at 18:40
  • $\begingroup$ @losif Pinelis I'm not sure I've understood you correctly, but what kind of shifting (wlog) do you mean? I have only one idea: if $P(x)=(x−x_1)(x−x_2)(x−x_3)(x−x_4)$ then with the replacement $x=u+\frac{x_1+x_2}{2}$ we will really have $P(x)=(u^2−a^2)(u−b)(u−c)$. But after finding $α,β,γ$ for $u$, we will see that they are not suitable for $x$. $\endgroup$ – Paul Ivanov May 15 at 10:17
  • $\begingroup$ @PaulIvanov : You will have to shift it back. Specifically, if $w=x-c=\frac{\alpha t+\beta}{\gamma t+1}$, then $x=w+c=\frac{(\alpha+c\gamma) t+(\beta+c)}{\gamma t+1}$. $\endgroup$ – Iosif Pinelis May 15 at 11:53
  • $\begingroup$ I have completely completed the answer, by adding the consideration of the previously remaining exceptional cases. $\endgroup$ – Iosif Pinelis May 15 at 16:33
  • $\begingroup$ @losif Pinelis, the problem is not that after reverse replacing $x$ can not be expressed as $\frac{\alpha' t+\beta'}{\gamma' t+1}$ with some $\alpha',\beta',\gamma'$. $\endgroup$ – Paul Ivanov 2 days ago

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