SUMO-6

Consider the function $f(x) = \left(\dfrac {x^2-10x+26}5\right)^{x^2-6x+5}$ $= \color{#3D99F6} \left(\dfrac {(x-3)(x-7)+5}5\right)^{\color{#D61F06}(x-1)(x-5)}$ $= \color{#3D99F6} g(x)^{\color{#D61F06}h(x)}$ .

We note that $f(x) = 1$ , when $\begin{cases} g(x) = 1 & \implies x=3, 7 \\ h(x) = 0 \text{ and } g(x) \ne 0 & \implies x = 1, 5 \end{cases}$

Therefore, $a+b+c+d = 1+3+5+7 = \boxed{16}$ .

For a solution to exist, we should have $x^2-6x+5=0 \implies (x-5)(x-1)=0 \implies x=5,1$ and $\frac{1}{5}(x^2-10x+26)=1\implies (x-3)(x-7)=0, x=3,7.$

If we look at $(x^2-10x+26)=(x-5)^2+1 > 0$ , it is always positive. For a positive $a$ , the function $f(y)=a^y$ has to be strictly increasing if $y=0 \text{ and } a^y=1.$

Thus the values for $x$ are $1,5,3 \text { and } 7 \implies 1+3+5+7=\boxed{16}.$