8. Find all integer values of the parameter $a$ for which the system $\left\{\begin{array}{l}x-2 y=y^{2}+2, \\ a x-2 y=y^{2}+x^{2}+0.25 a^{2}\end{array}\right.$. has at least one solution. In your answer, specify the sum of the found values of the parameter $a$.
Solution: Transform the system

$$
\left\{\begin{array} { l } 
{ x - 1 = ( y + 1 ) ^ { 2 } } \\
{ ( y + 1 ) ^ { 2 } + ( x - 0.5 a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x-1=(y+1)^{2} \\
x-2+(x-0.5 a)^{2}=0
\end{array}\right.\right.
$$

Consider the second equation of the system

$x^{2}+x(1-a)+\left(a^{2}-8\right) / 4=0, \quad D=(a-1)^{2}-\left(a^{2}-8\right)=9-2 a$

A solution exists when $a \leq 9 / 2, x=\frac{a-1 \pm \sqrt{9-2 a}}{2}$, and $x \geq 1$. For the solution to exist, the following conditions must be met:

$$
\left\{\begin{array} { l } 
{ D = 9 - 2 a \geq 0 } \\
{ \left[ \begin{array} { l } 
{ f ( 1 ) = ( a ^ { 2 } - 4 a ) / 4 > 0 } \\
{ \frac { a - 1 } { 2 } > 1 }
\end{array} \right. } \\
{ f ( 1 ) = ( a ^ { 2 } - 4 a ) / 4 \leq 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a \leq 9 / 2 \\
{\left[\begin{array}{l}
a(a-4)>0 \\
a-1>2 \\
a(a-4) \leq 0
\end{array}\right.}
\end{array} \Rightarrow a \in[0,9 / 2]\right.\right.
$$

Summing the integer values of the parameter, we get $0+1+2+3+4=10$.

Answer: 10.