Is it possible to find the sum of all integer values that $x$ can take? Is it possible to find the sum of all integer values that $x$ can take? In:
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$

Make the substitution $x=t^2+1$
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
$$\Rightarrow\sqrt{(t^2+1)+3-4\sqrt{(t^2+1)-1}}+\sqrt{(t^2+1)+8-6\sqrt{(t^2+1)-1}}=1$$
$$\Rightarrow\sqrt{t^2+1+3-4|t|}+\sqrt{t^2+1+8-6|t|}=1$$
$$\Rightarrow\sqrt{|t|^2-4|t|+4}+\sqrt{|t|^2-6|t|+9}=1$$
Make the substituion $|t|=a$
$$\Rightarrow\sqrt{(a-2)^2}+\sqrt{(a-3)^2}=1$$
$$\Rightarrow |a-2|+|a-3|=1$$
$$\Rightarrow |3-a|+|a-2|=1$$
We can see $(3-a)+(a-2)=1$ And we know that
$$|a|+|b|\geq|a+b|$$
And equality holds when $a$ and $b$ have like sign. So we have,
$$(3-a)(a-2)\geq 0$$
$$\Rightarrow (a-3)(a-2)\leq 0$$
$$\Rightarrow a\in [2,3] $$
Reverting back to $t$
$$\Rightarrow t\in [2,3]\cup [-3,-2] $$
Since $t=\sqrt{a-1}$, we have
$$x\in [5,10] $$
