Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-8 \\
y^{2}=(z-x)^{2}-20 \\
z^{2}=(x-y)^{2}+40
\end{array}\right.
$$
Answer: $(-1,-4,-7),(1,4,7)$.

Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:

$$
\begin{cases}(x-y+z)(x+y-z) & =-8 \\ (y-z+x)(y+z-x) & =-20 \\ (z-x+y)(z+x-y) & =40\end{cases}
$$

Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then

$$
\begin{cases}Y Z & =-8 \\ Z X & =-20 \\ X Y & =40\end{cases}
$$

Multiplying all the resulting equations, we have $(X Y Z)^{2}=8 \cdot 20 \cdot 40$, from which $X Y Z=80$ or $X Y Z=$ -80. Consider the case $X Y Z=80$. In this case, $X=(X Y Z) /(Y Z)=-10, Y=-4, Z=2$; then $x=\frac{Y+Z}{2}=$ $-1, y=-4, z=-7$. The second case is handled similarly and in it $x=1, y=4, z=7$.