<s>[INST] Super hard system of equations 
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...
 [/INST] 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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