jmc

Let $P(x)=x^2+ax+b$ and $Q(x)=x^2+ax-b.$ We seek $a$ and $b$ so that $Q(P(x))$ has a single real repeated root.

Let the roots of $Q(x)$ be $r_1$ and $r_2.$ Then the roots of $Q(P(x))$ are the roots of the equations $P(x)=r_1$ and $P(x)=r_2.$ Therefore, $Q(x)$ must have a repeated root, which means its discriminant must be 0. This gives us $a^2+4b=0.$ The repeated root of $Q(x)=x^2+ax-b$ is then $-\frac{a}{2}.$

Then, the equation $P(x)=-\frac{a}{2}$ must also have a repeated root. Writing out the equation, we get $x^2+ax+b=-\frac{a}{2},$ or $$x^2+ax+\frac{a}{2}+b=0.$$Again, the discriminant must be 0, so $a^2-2a-4b=0.$ We know $4b=-a^2,$ so $$2a^2-2a=2a(a-1)=0.$$Hence, $a=0$ or $a=1.$ If $a=0,$ then $b=0.$ If $a=1,$ then $b=-\frac{1}{4}.$ Thus, the solutions $(a,b)$ are $(0,0)$ and $\left(1,-\frac{1}{4}\right),$ and the final answer is $0+0+1-\frac{1}{4}=\boxed{\frac{3}{4}}.$