Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-3 \\
y^{2}=(z-x)^{2}-7 \\
z^{2}=(x-y)^{2}+21
\end{array}\right.
$$
Answer: $(-1,-3,-5),(1,3,5)$.

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

$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=-3 \\
(y-z+x)(y+z-x)=-7 \\
(z-x+y)(z+x-y)=21
\end{array}\right.
$$

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

$$
\left\{\begin{aligned}
Y Z & =-3 \\
Z X & =-7 \\
X Y & =21
\end{aligned}\right.
$$

Multiplying all the obtained equations, we have $(X Y Z)^{2}=3 \cdot 7 \cdot 21$, from which $X Y Z=21$ or $X Y Z=-21$. Consider the case $X Y Z=21$. In this case, $X=(X Y Z) /(Y Z)=-7, Y=-3, Z=1$; then $x=\frac{Y+Z}{2}=-1, y=-3, z=-5$. The second case is considered similarly, and in it $x=1, y=3, z=5$.