# Transformation of a fourth degree polynomial

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.


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.

• @PaulIvanov : I have now given a completely different solution. May 14, 2019 at 18:40
• @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 $x$. May 15, 2019 at 10:17
• @PaulIvanov : You will have to shift it back. Specifically, if $w=x-c=\frac{\alpha t+\beta}{\gamma t+1}$, then $x=w+c=\frac{(\alpha+c\gamma) t+(\beta+c)}{\gamma t+1}$. May 15, 2019 at 11:53
• I have completely completed the answer, by adding the consideration of the previously remaining exceptional cases. May 15, 2019 at 16:33
• @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 $\alpha',\beta',\gamma'$. May 17, 2019 at 10:24

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.

• 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 May 13, 2019 at 18:31
• 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 $|A|<2$. May 14, 2019 at 13:57
• @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). May 14, 2019 at 14:19
• 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? May 14, 2019 at 18:46
• Also, as seen from my answer, there are exceptions when the desired representation is not possible. Why don't they appear in your answer? May 14, 2019 at 18:48