66. $\{x+y=10$,

$\left\{x^{2}+y^{2}=68\right.$
66. Diophantus' Solution.

$$
\frac{x+y}{2}=5
$$

Let

$$
\frac{x-y}{2}=d
$$

Adding gives $x=5+d$, subtracting gives $y=5-d .(5+d)^{2}+$ $+(5-d)^{2}=68 ; 50+2 d^{2}=68 ; 2 d^{2}=18 ; d=3 ; x=8 ; y=2$.

Regular solution:

$$
\begin{aligned}
&-x^{2}+y^{2}+2 x y=100 \\
&-x^{2}+y^{2}=68 \\
& 2 x y=32 \\
& x^{2}+y^{2}-2 x y=(x-y)^{2}=36 ; x-y=6
\end{aligned}
$$

Then combine with $x+y=10$.