Mfrhtyj

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...

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 $tneq 0$.
(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.)

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:
$$2abcd-ab(c+d)^2-cd(a+b)^2=4tag{1*}$$
From $(3)$, we have:
$$begin{align}
(a^3+b^3)(c^3+d^3)&=color{red}{(a+b)}((a+b)^2-3ab)color{red}{(c+d)}((c+d)^2-3cd)=7\
&=((a+b)^2-3ab)((c+d)^2-3cd)=7\
&qquadvdots\
&=3abcd-ab(c+d)^2-cd(a+b)^2=2tag{2*}
end{align}$$
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:
$$begin{align}
left((a+b)^4+2a^2b^2-4ab(a+b)^2right)left((c+d)^4+2c^2d^2-4cd(c+d)^2right)&=25
end{align}$$
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:
$$color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}+5ab(c+d)^2-10cd(a+b)^2+16=12tag{3*}$$
However, you can see that the text in $color{red}{text{red}}$ looks very close to the square of two sums:
$$color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}=(ab(c+d)^2+cd(a+b)^2)^2-2color{blue}{abcd(a+b)^2(c+d)^2}$$
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.

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.

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 :

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

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 :

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

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 $alpha=P_1P_2=-2.$

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
$$left{a= frac{2}{d},b= frac{sqrt{2}}{d},c=
-frac{d}{sqrt{2}}right},left{a= frac{sqrt{2}}{d},b= frac{2}{d},c=
-frac{d}{sqrt{2}}right},left{a= frac{2}{d},b=
-frac{sqrt{2}}{d},c= frac{d}{sqrt{2}}right},left{a=
-frac{sqrt{2}}{d},b= frac{2}{d},c= frac{d}{sqrt{2}}right},left{a=
-frac{1}{d},b= -frac{sqrt{2}}{d},c= -sqrt{2} dright},left{a=
-frac{sqrt{2}}{d},b= -frac{1}{d},c= -sqrt{2} dright},left{a=
-frac{1}{d},b= frac{sqrt{2}}{d},c= sqrt{2} dright},left{a=
frac{sqrt{2}}{d},b= -frac{1}{d},c= sqrt{2} dright}$$

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