<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] 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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