Solve $\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $ 

Solve $$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$


Preface; I think there should be an algebraic method of solving this equation for $x$ since graphing these two graphs 
We get whole number solutions such as $x=0,2,3$
So I think there is someway of manipulating this equation into a disguised quadratic somehow!
So my attempt is this: 
$$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$
Let $u=x^2-5x+8$ and $u-2=x^2-5x+6$ which means we can rewrite our equation as 
$$\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x + \left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x = 2^{\frac{x+4}{4}} $$
Squaring both sides we get 
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x + 2\left(\sqrt{\sqrt{u}+\sqrt{u-2}} \right)^x\left(\sqrt{\sqrt{u}-\sqrt{u-2}} \right)^x+ \left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+4}{2}}$$
$$\left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x+ 2(\sqrt{2})^x+  = 2^{\frac{x+4}{2}}$$
Now some little algebra
$2^{\frac{x+4}{2}}-2(\sqrt{2})^x=2^{\frac{x}{2}} \cdot 2^2 - 2 \cdot 2^{\frac{x}{2}}=2^{\frac{x}{2}} \cdot 2=2^{\frac{x+2}{2}}$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^x +\left( \sqrt{u} - \sqrt{u-2}\right)^x = 2^{\frac{x+2}{2}} $$
Square both sides again 
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2\left( \sqrt{u} + \sqrt{u-2}\right)^x\left( \sqrt{u} - \sqrt{u-2}\right)^x   = 2^{x+2} $$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}+2(2^x)    = 2^{x+2} $$
$$ \left( \sqrt{u} + \sqrt{u-2}\right)^{2x} +\left( \sqrt{u} - \sqrt{u-2}\right)^{2x}  = 2^{x+1} $$
Now I've hit a roadblock.. :(

HINT:
Replace $\sqrt u-\sqrt{u-2}=\dfrac2{\sqrt u+\sqrt{u-2}}$ to form a Quadratic equation in  $$\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}$$ which is
$$\left(\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}-1\right)^2=0$$
