<s>[INST] A function with domain that is asking for sum of a+b+c+d Let $f$ be a function such that $$ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $$for $x > 1$. In that domain, $f(x)$ has the form $\frac{a+\sqrt{cx+d}}{b},$ where $a,b,c,d$ are integers and $a,b$ are relatively prime. Find $a+b+c+d.$
 [/INST] Let $y = f(x)$. Our first goal is to find all pairs of real numbers $(x,y)$ that satisfy
[\sqrt{x - \sqrt{x + y}} = y.]We observe that $y$ must be nonnegative. Squaring both sides, we get
[x - \sqrt{x + y} = y^2,]so
[\sqrt{x + y} = x - y^2.]Squaring both sides again, we get
[x + y = x^2 - 2xy^2 + y^4,]so
[y^4 - 2xy^2 - y + x^2 - x = 0.]This is a quartic equation in $y$, which has no obvious solutions. However, we can re-write it as a quadratic equation in $x$, which we can solve:
[x^2 - (2y^2 + 1)x + y^4 - y = 0.]By the quadratic formula,
\begin{align*}
x &= \frac{(2y^2 + 1) \pm \sqrt{(2y^2 + 1)^2 - 4(y^4 - y)}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{4y^4 + 4y^2 + 1 - 4y^4 + 4y}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{4y^2 + 4y + 1}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{(2y + 1)^2}}{2} \\
&= \frac{2y^2 + 1 \pm (2y + 1)}{2},
\end{align*}so
[x = \frac{2y^2 + 1 + 2y + 1}{2} = y^2 + y + 1,]or
[x = \frac{2y^2 + 1 - (2y + 1)}{2} = y^2 - y.]We consider these solutions separately.
Case 1: $x = y^2 + y + 1$.
In this case,
[x + y = y^2 + 2y + 1 = (y + 1)^2,]so
[\sqrt{x + y} = |y + 1|.]We know that $y \ge 0$, so $y + 1 \ge 1$. Hence, $\sqrt{x + y} = y + 1$. Then
[x - \sqrt{x + y} = y^2 + y + 1 - (y + 1) = y^2,]so
[\sqrt{x - \sqrt{x + y}} = \sqrt{y^2} = |y| = y.]Thus, if $x = y^2 + y + 1$, then $(x,y)$ satisfies the given equation. We also know that $y \ge 0$, so $x \ge 1$. From the equation $x = y^2 + y + 1$, we have that $y^2 + y + 1 - x = 0$, so by the quadratic equation,
[y = \frac{-1 \pm \sqrt{1 - 4(1 - x)}}{2} = \frac{-1 \pm \sqrt{4x - 3}}{2}.]Since $y \ge 0$, we must take the root with the plus sign. Hence,
[y = \frac{-1 + \sqrt{4x - 3}}{2}.]
Case 2: $x = y^2 - y$.
Recall above that we also found $\sqrt{x+y}=x-y^2.$ Substituting $x=y^2-y$ gives
[\sqrt{y^2}=-y,]which is only satisfied when $y=0$, in which case $x = 0$, but we are given that $x > 0$.
We conclude that the function $f(x)$ is given by
$$f(x) = \frac{-1 + \sqrt{4x - 3}}{2}.$$This gives $a = -1, b = 2, c = 4, d = -3,$ or $a+b+c+d = \boxed{ 2 }.$
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