Answer
$(2,2), (-2,-2)$
Work Step by Step
1. Let the two numbers be $x, y$, based on the given conditions, we can get:
$\begin{cases} xy=4 \\ x^2+y^2=8 \end{cases}$
2. The 1st equation gives $y=\frac{4}{x}$, use it in the 2nd equation to get $x^2+16/x^2=8$ or $x^4-8x^2+16=0$ or $(x^2-4)^2=0$ thus $x=\pm2$
3. Use the 1st equation again to get the solution(s) $(2,2), (-2,-2)$
