What kind of Sorcery is this ?

a^2-b^2=(a+b)(a-b)=2\tag{1}

We'll use this fact multiple times. Now square both sides of the initial equation:

$(a+b)^x+(a-b)^x+2\cdot 2^{\frac{x}{2}}=(\sqrt{2})^{x+4}=4\cdot 2^{\frac{x}{2}}$

$\implies (a+b)^x+(a-b)^x=2\cdot 2^{\frac{x}{2}}$

Now square both sides again.

$((a+b)^2)^x+((a-b)^2)^x+2\cdot 2^x=4\cdot 2^x$

$\implies ((a+b)^2)^x+((a-b)^2)^x=2\cdot 2^x$

\stackrel{:2^x}\implies \left(\frac{(a+b)^2}{2} \right)^x+\left(\frac{(a-b)^2}{2} \right)^x=2

\stackrel{(1)}\implies \left(\frac{(a+b)^2}{2} \right)^x+ \left(\frac{2}{(a+b)^2} \right)^x=2

The 2 terms on the LHS are positive, hence:

$\color{royalblue}{\textbf{AM-GM:}} \left(\frac{(a+b)^2}{2} \right)^x+ \left(\frac{2}{(a+b)^2} \right)^x \ge 2\sqrt{\left(\frac{(a+b)^2}{2} \right)^x \left(\frac{2}{(a+b)^2} \right)^x}= 2$

Hence the LHS is equal to its minimum value, which is achieved if and only if

$\left(\frac{(a+b)^2}{2} \right)^x=\left(\frac{2}{(a+b)^2} \right)^x$

$\implies (a+b)^{4x}=(\sqrt{2})^{4x}$

If $x=0$ , then this equation always holds. After checking $x=0$ on the initial conditions, we see it works. Thus $\boxed{x_1=0}$ is one solution.

If $x\neq 0$ , then

a+b=\sqrt{2}\stackrel{(1)}\implies a-b=\sqrt{2}

$\begin{cases}a+b=\sqrt{2}\\a-b=\sqrt{2}\end{cases}\implies \begin{cases}a=\sqrt{2}\\b=0\end{cases}$

$\implies \sqrt{(x-1)(x-4)}=0\implies \begin{cases}\boxed{x_2=1}\\\text{or}\\\boxed{x_3=4}\end{cases}$

We see that these 2 solutions satisfy all the conditions. Thus summing all the boxed solutions gives:

$\color{royalblue}{\textbf{answer}}=x_1+x_2+x_3-1=0+1+4-1=\boxed{4}. \square$

${ (a+b) }^{ \frac { x }{ 2 } }+{ (a-b) }^{ \frac { x }{ 2 } }={ 2 }^{ \frac { x+4 }{ 4 } }\\ Squaring\quad both\quad sides\\ { (a+b) }^{ x }+{ (a-b) }^{ x }+2{ ({ a }^{ 2 }{ -b }^{ 2 }) }^{ \frac { x }{ 2 } }={ 2 }^{ \frac { x+4 }{ 2 } }\\ From\quad the\quad given\quad equations,\quad ({ a }^{ 2 }{ -b }^{ 2 })=2\\ { (a+b) }^{ x }+{ (a-b) }^{ x }+2{ (2) }^{ \frac { x }{ 2 } }={ 2 }^{ \frac { x+4 }{ 2 } }\\ { (a+b) }^{ x }+{ (a-b) }^{ x }+{ 2 }^{ \frac { x+2 }{ 2 } }={ 2 }^{ \frac { x+4 }{ 2 } }\\ { (a+b) }^{ x }+{ (a-b) }^{ x }={ 2 }^{ \frac { x+4 }{ 2 } }-{ 2 }^{ \frac { x+2 }{ 2 } }\\ { (a+b) }^{ x }+{ { (a-b) }^{ x } }={ 2 }^{ \frac { x }{ 2 } }(2)\\ { (a+b) }^{ x }+{ (a-b) }^{ x }-{ 2 }^{ \frac { x+2 }{ 2 } }=0\\ { (a+b) }^{ x }+{ (a-b) }^{ x }-2{ (2) }^{ \frac { x }{ 2 } }=0\\ { [{ (a+b) }^{ \frac { x }{ 2 } }-{ (a-b) }^{ \frac { x }{ 2 } }] }^{ 2 }=0\\ { (a+b) }^{ \frac { x }{ 2 } }={ (a-b) }^{ \frac { x }{ 2 } }\\ a+b=a-b\\ b=-b\\ b=0\\ Given:\\ b=\sqrt { { x }^{ 2 }-5x+4 } \\ b=\sqrt { { (x }-1)(x-4) } \\ b=0\quad when\quad x=1\quad or\quad 4\\ Sum=4+1=5\\ Answer(predecessor)=\boxed { 4 }$

(a+b)x/2 + (a-b)x/2 = 2[(x+4)/4]
2ax = x+4 and a=[(x-3)(x-2)]^1/2
4x^4 - 20x^3 + 24x^2 = x^2 + 8x + 16 x^4 - 5x^3 + (23x^2)/4 - 8x - 16 = 0 = (x-p)(x-q)(x-r)(x-s)
All the values of x are p,q,r and s
-(p+q+r+s) = -5 (Vieta)
So the answer is 4