Strange System of Equations

First, we rearrange the first equation to get $y^2-x^2+2x+2y=\sqrt{x}$ which factors into $(y+x)(y-x+2)=\sqrt{x}$ or $y-x+2=\dfrac{\sqrt{x}}{x+y}$

Now looking at the second equation, we see that $x+y=\dfrac{1}{2\sqrt{x}}$ Substituting this back into the first equation gives $y-x+2=\dfrac{\sqrt{x}}{\dfrac{1}{2\sqrt{x}}}=2x$

Now subtracting the second equation from this new equation gives $2-2x=2x-\dfrac{1}{2\sqrt{x}}$ or $4x-\dfrac{1}{2\sqrt{x}}=2$

Multiplying both sides by $\sqrt{x}$ gives $4x\sqrt{x}-\dfrac{1}{2}=2\sqrt{x}$ or $8x\sqrt{x}-4\sqrt{x}-1=0$

Substituting $z=\sqrt{x}$ , we have $8z^3-4z-1=0$

Upon inspection with the Rational Root Theorem, we see that $-\dfrac{1}{2}$ is a root, thus $\left(x+\dfrac{1}{2}\right)$ is a factor. Using synthetic division, we see that $8z^3-4z-1=2\left(z+\dfrac{1}{2}\right)(4z^2-2z-1)$

The roots of $4z^2-2z-1$ , using the quadratic equation, are $z=\dfrac{1\pm \sqrt{5}}{4}$

However, see that since $z=\sqrt{x}$ , we must have that $z$ is positive. Since both $z=-\dfrac{1}{2}$ and $z=\dfrac{1-\sqrt{5}}{4}$ are negative, we must have $z=\dfrac{1+\sqrt{5}}{4}$ .

Thus, $\sqrt{x}=\dfrac{1+\sqrt{5}}{4}$ or $x=\dfrac{3+\sqrt{5}}{8}$

Plugging $\sqrt{x}$ and $x$ back into the second equation, we see that $\begin{aligned}\dfrac{3+\sqrt{5}}{8}+y&=\dfrac{1}{2\left(\dfrac{1+\sqrt{5}}{4}\right)}\\ \dfrac{3+\sqrt{5}}{8}+y&=\dfrac{2}{1+\sqrt{5}}\\ \dfrac{3+\sqrt{5}}{8}+y&=\dfrac{\sqrt{5}-1}{2}\\ y&= \dfrac{\sqrt{5}-1}{2}-\dfrac{3+\sqrt{5}}{8}\\ &= \dfrac{3\sqrt{5}-7}{8}\end{aligned}$

Thus, $\begin{aligned}xy&=\left(\dfrac{3+\sqrt{5}}{8}\right)\left(\dfrac{3\sqrt{5}-7}{8}\right)\\ &= \dfrac{-3+\sqrt{5}}{32}\end{aligned}$

Thus our final answer is $-3+5+32=\boxed{34}$ and we are done.

For me,

a = x + y

b = y - x

$\Rightarrow$ $y^2 + 2y = x^2 -2x+ \sqrt{x} <=> a^2 (b+2)= \frac{1}{2} (1)$

$\Rightarrow$ $x + y = \frac{1}{2\sqrt{x}} <=> a^2(a - b) = \frac{1}{2} (2)$

$(1) - (2) => x = \frac{3 + \sqrt{5}}{8} ; y = \frac{3\sqrt{5} -7}{8}$

$\Rightarrow$ $xy = \frac{-3 + \sqrt{5}}{32}$

$\Rightarrow$
$a + b + c = \boxed{34}$