8. A number plus 79 becomes a perfect square, and when this number plus 204 is added, it becomes another perfect square. Then this number is $\qquad$ .
$$
\begin{array}{c}
204=y^{2}, \text { where } x \text { and } y \text { are non-negative integers. Thus, } \\
y^{2}-x^{2}=125=5^{3}, \text { i.e., } (y+x)(y-x)=5^{3}. \\
\therefore \quad\left\{\begin{array} { l } 
{ y + x = 125, } \\
{ y - x = 1; }
\end{array} \text { or } \left\{\begin{array} { l } 
{ y + x = 25, } \\
{ y - x = 5. }
\end{array} \text { Solving, we get } \left\{\begin{array} { l } 
{ x = 62, } \\
{ y = 63; }
\end{array} \text { or } \left\{\begin{array}{l}
x=10, \\
y=15. 
\end{array}\right.\right.\right. \\
\therefore \quad a=62^{2}-79=3765,
\end{array}
$$

or $a=10^{2}-79=21$.