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Solve the system of equation for real numbers

\begin{split} (a+b) &(c+d) &= 1 & \qquad (1)\\ (a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\ (a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\ (a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\ \end{split}


First I used the identity $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$ Use this identity to (4) too and simplify (3), we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$ And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$ But got stuck...

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    $\begingroup$ I would use that $$a^3+b^3=(a+b)(a^2+b^2-ab)$$ etc $\endgroup$ – Dr. Sonnhard Graubner Jan 6 at 13:23
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    $\begingroup$ Your system is a very interesting (and puzzling!). Could you say where you found it ? $\endgroup$ – Jean Marie Jan 7 at 7:24
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Hint: $ac=x, bc=y, ad=u, bd=v$, then the equations are

$x+y+u+v=1$

$x^2+y^2+u^2+v^2=9$

$x^3+y^3+u^3+v^3=7$

$x^4+y^4+u^4+v^4=25$

Use Newton-Girard to compute the elementary polynomials. Then you have the polynomial $P(z)= (z-x)(z-y)(z-u)(z-v)$ with variable $z$. Solve the quartic equation $P(z)=0$, and there you have the values $x,y,u,v$ in some order. Note that not in any order: $xv=yu$ must be true, see the definition of these variables.

Of course, once you have $x,y,u,v$, it is easy to compute $a,b,c,d$.

P.S. By this way we can get: $$\{x,y,u,v\}=\{-1,2,\sqrt2,-\sqrt2\},$$ which gives $abcd=-2.$ Up to symmetry, the solution is $(a,b,c,d)= (t, -\sqrt{2}t, -\frac{1}{t}, -\frac{\sqrt{2}}{t})$ for any $t\neq 0$. (By up to symmetry, I mean you can switch $a$ and $b$, you can switch $c$ and $d$, and you can switch the pair $(a,b)$ with $(c,d)$, so there are $8$ symmetries.)

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    $\begingroup$ I solved your system. What is the rest? $\endgroup$ – Michael Rozenberg Jan 6 at 14:10
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    $\begingroup$ Yes, I see now. We can kill it. $\endgroup$ – Michael Rozenberg Jan 6 at 14:18
  • $\begingroup$ I am sorry but I don't find your derivation very clear. Besides, a) the solution you propose doesn't work : from the four equations $(1) ac=-1=x, (2) bc=2=y, (3) ad=\sqrt{2}=u, (4) bd=-\sqrt{2}=v$, if I divide (1) by (3), I get $c/d=-1/\sqrt{2}$, whereas, if I divide (2) by (3), I get $c/d=\sqrt{2}$, which is contradictory. b) See the infinite family of solutions provided by Claude. $\endgroup$ – Jean Marie Jan 6 at 17:05
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    $\begingroup$ @JeanMarie I am sorry if you don't find it clear, but could you be more specific? I checked, and it seems to me that the solution (given by @MichaelRozenberg) is correct for $x,y,u,v$. From that it should be a routine taks to solve the original system of equations. You made a mistake though: as I said, you cannot randomly pick which root is which variable. It should be $ab=-1, cd=2, ac=\sqrt{2}, bd=-\sqrt{2}$, or one of the other seven symmetrical choices. Keep in mind that $xv=yu$. $\endgroup$ – A. Pongrácz Jan 6 at 19:04
  • $\begingroup$ I realize that I have misunderstood your set equality $$\{x,y,u,v\}=\{-1,2,\sqrt2,-\sqrt2\}$$ that I have interpreted as a t-uple equality $$(x,y,u,v)=(-1,2,\sqrt2,-\sqrt2)$$... But what I find not evident to understand is a) the (clever !) transformation of the initial system into yours b) what is exactly the set of solutions, being understood that it depends (at least) on an arbitrary constant : why/how that ? $\endgroup$ – Jean Marie Jan 6 at 19:14
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Not really complete, but an interesting result using simple algebraic manipulations.


Write: $$\begin{align} a^2+b^2&=(a+b)^2-2ab\\ a^3+b^3&=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)\\ a^4+b^4&=((a^2)^2+(b^2)^2)=\cdots=(a+b)^4+2a^2b^2-4ab(a+b)^2\\ \vdots \end{align}$$ From $(2)$, we have: $$\begin{align} (a^2+b^2)(c^2+d^2)&=((a+b)^2-2ab)((c+d)^2-2cd)=9\\ &=\color{red}{(a+b)^2(c+d)^2}-2ab(c+d)^2-2cd(a+b)^2-4abcd=9 \end{align}$$ But from $(1)$, we know that $(a+b)(c+d)=1$, then the text in $\color{red}{\text{red}}$ is also equal to $1$, so the result above becomes: $$2abcd-ab(c+d)^2-cd(a+b)^2=4\tag{1*}$$ From $(3)$, we have: $$\begin{align} (a^3+b^3)(c^3+d^3)&=\color{red}{(a+b)}((a+b)^2-3ab)\color{red}{(c+d)}((c+d)^2-3cd)=7\\ &=((a+b)^2-3ab)((c+d)^2-3cd)=7\\ &\qquad\vdots\\ &=3abcd-ab(c+d)^2-cd(a+b)^2=2\tag{2*} \end{align}$$ Adding $(1*)$ and $(2*)$, we get $abcd=-2$ and that $\color{pink}{ab(c+d)^2+cd(a+b)^2=-8}$.


Now $(4)$ is really tricky, but you can write it as: $$\begin{align} \left((a+b)^4+2a^2b^2-4ab(a+b)^2\right)\left((c+d)^4+2c^2d^2-4cd(c+d)^2\right)&=25 \end{align}$$ Expanding, and we can eliminate $(a+b)^4(c+d)^4$ since it is equal to $1$. Then we have: $$\begin{align} a^2b^2(c+d)^4-2ab(c+d)^2+c^2d^2(a+b)^4+2(abcd)^2-4abc^2d^2(a+b)^2-2cd(a+b)^2-4a^2b^2cd(c+d)^2-4abcd=12 \end{align}$$ Using the fact that $abcd=-2$, then we can shorten the equation above into: $$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}+5ab(c+d)^2-10cd(a+b)^2+16=12\tag{3*}$$ However, you can see that the text in $\color{red}{\text{red}}$ looks very close to the square of two sums: $$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}=(ab(c+d)^2+cd(a+b)^2)^2-2\color{blue}{abcd(a+b)^2(c+d)^2}$$ However we already know the value of the part in $\color{blue}{\text{blue}}$ to be $-2 \cdot 1$. Now we can write $(3*)$ as: $$(ab(c+d)^2+cd(a+b)^2)^2+6ab(c+d)^2-10cd(a+b)^2=-8$$


From here, you can substitute $x=ab(c+d)^2$ and $y=cd(a+b)^2$, which gives two systems of equation: $$(x+y)^2+6x-10y=-8\\ x+y=-8$$ This has one solution: $$x=-\frac{19}2\,\,y=\frac32$$


You can try working from here.

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  • $\begingroup$ You and I have been working on the same tracks : obtaining values for expressions involving $a+b$ and $ab$ (such as my $\alpha$ and $\beta$) that provide equations. Afterwards, I would say that it's a matter of luck to find the right equations... $\endgroup$ – Jean Marie Jan 7 at 7:54
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We are going to show that $(a,b)$ belongs to one of the two lines with equations $b=\sqrt{a}$ and $b=\frac{1}{\sqrt{a}}$ as displayed on the following figure. It will give the answer, due to the symmetry of the system of equations with respect to the group of variables $(a,b)$ vs. $(c,d)$. Moreover, we will establish (see (*) at the bottom) that the last equation is superfluous.

enter image description here

Here is the explanation :

Let :

$$S_1:=a+b, \ \ S_2:=c+d, \ \ P_1:=ab, \ \ P_2:=cd$$

The system constituted by the first three equations can be written, with these variables, using classical transformations :

$$\begin{cases} (A) \ &S_1S_2&=&1& \ &\\ (B) \ &(S_1^2-2P_1)(S_2^2-2P_2)&=&9 & \ \implies \ & (C) \ 1-2(P_1S_2^2+P_2S_1^2)+4(P_1P_2)=9\\ (D) \ &(S_1^3-2P_1S_1)(S_2^3-2P_2S_2)&=&7 & \ \implies \ & (E) \ 1-3S_1S_2(P_1S_2^2+P_2S_1^2)+9(P_1P_2)=7. \end{cases}$$

(equations (C) and (E) are obtained by expansion of (B) and (D) resp., using relationship (A)).

Setting

$$\alpha := P_1P_2 \ \text{and} \ \beta := P_1S_2^2+P_2S_1^2,$$

equations (C) and (E) become :

$$\begin{cases} (C) & \ 2\alpha-\beta&=&4\\ (E) & \ 3\alpha-\beta&=&2 \end{cases} \ \ \implies \ \ \alpha=-2 \ \text{and} \ \beta=-8.$$

Using the fact that $S_1S_2=1$ and $\alpha=P_1P_2=-2$, equation $\beta=-8$ becomes :

$$P_1 \frac{1}{S_1^2} - \frac{2}{P_1}S_1^2 = -8$$

i.e.,

$$(F) \ \ \ \ P_1^2 + 8 P_1S_1^2 - 2 S_1^4 =0,$$

which can be considered as a quadratic equation in variable $P_1$ giving two solutions. Due to classical condition

$$(a+b)^2 \geq 2ab \ \iff \ S_1^2 \geq 2P_1,$$

only one of these solutions is eligible :

$$P_1=(-4+3\sqrt{2})S_1^2 \ \ \ \iff \ \ \ ab=(-4+3\sqrt{2})(a+b)^2 \ \ \ \iff \ \ \ (b-\sqrt{2}a)(b-\frac{\sqrt{2}}{2}a)=0$$

whence the result corresponding to the figure.

The parametric equations of the two lines are

$$(a,b)=(p,p \sqrt{2}) \ \ \text{and} \ \ (a,b)=(p,p \frac{\sqrt{2}}{2}), \ \ \text{for any} \ \ p \neq 0$$

Due to the symmetry of equations, we have as well, for any $q \neq 0$ :

$$(c,d)=(q,q \sqrt{2}) \ \ \text{and} \ \ (c,d)=(q,q \frac{\sqrt{2}}{2}).$$

A quick glance at any of the four equations show that necessarily $q=\frac{1}{p}$. We find back in this way all the solutions given by @Claude Leibovici and @A. Pongrácz .


(*) In fact, the fourth equation is a consequence of the first three. Here is why :

First of all, relationship (F) is equivalent to :

$$(G) \ \ \ \ S_1^4=\frac12P_1^2+4P_1S_1^2.$$

As the fourth equation can be written :

$$(H) \ \ \ \ (S_1^4+2P_1^2-4P_1S_1^2)(S_2^4+2P_2^2-4P_2S_2^2)=25,$$

using (G) in (H), we get :

$$\frac52P_1^2 \frac52P_2^2=25,$$

which is a tautology due to the fact that $\alpha=P_1P_2=-2.$

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This is not an answer but it is too long for a comment.

Looking at this system of equations, I had a very strange feeling (which I cannot explain).

Using a CAS, I solved equations $(1)$, $(2)$, $(3)$ for $a,b,c$ as functions of $d$ and obtained $8$ solutions which are listed below $$\left\{a= \frac{2}{d},b= \frac{\sqrt{2}}{d},c= -\frac{d}{\sqrt{2}}\right\},\left\{a= \frac{\sqrt{2}}{d},b= \frac{2}{d},c= -\frac{d}{\sqrt{2}}\right\},\left\{a= \frac{2}{d},b= -\frac{\sqrt{2}}{d},c= \frac{d}{\sqrt{2}}\right\},\left\{a= -\frac{\sqrt{2}}{d},b= \frac{2}{d},c= \frac{d}{\sqrt{2}}\right\},\left\{a= -\frac{1}{d},b= -\frac{\sqrt{2}}{d},c= -\sqrt{2} d\right\},\left\{a= -\frac{\sqrt{2}}{d},b= -\frac{1}{d},c= -\sqrt{2} d\right\},\left\{a= -\frac{1}{d},b= \frac{\sqrt{2}}{d},c= \sqrt{2} d\right\},\left\{a= \frac{\sqrt{2}}{d},b= -\frac{1}{d},c= \sqrt{2} d\right\}$$

The problem is that, replacing in $(4)$ any of these solutions the resulting equation is $25=25$ !

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    $\begingroup$ It is not a problem, Claude: you are perfectly right : if $(a,b,c,d)$ is a solution, $(a/k, b/k , ck, dk)$ is clearly a solution as well... Besides, bonne année 2019 ! $\endgroup$ – Jean Marie Jan 6 at 16:36
  • $\begingroup$ @JeanMarie. Good to see you here ! All my best wishes. Thanks for the comment. $\endgroup$ – Claude Leibovici Jan 6 at 17:03
  • $\begingroup$ This issue is puzzling : the fact that we have an infinite set of solutions means a kind of (algebraic) dependency of this set of equations. But, how to find it, and, in the case such a dependency is found, how can we exploit it ? $\endgroup$ – Jean Marie Jan 6 at 17:13

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