jmc

We can write $$\frac{a^2+b^2}{a-b}=\frac{a^2+b^2-2ab+16}{a-b}=\frac{(a-b{)}^2+16}{a-b}=a-b+\frac{16}{a-b}.$$By AM-GM, $$a-b+\frac{16}{a-b}\ge 2\sqrt{(a-b)\cdot \frac{16}{a-b}}=8.$$Equality occurs when $a-b=4$ and $ab=8.$ We can solve these equations to find $a=2\sqrt{3}+2$ and $b=2\sqrt{3}-2.$ Thus, the minimum value is $\boxed{8}.$