Don't write all the possibilities, part 2

What I noticed is that if $n$ is a whole number, and the value of $x$ is in the form $n(n+1) +1$ , then the value of $y$ will simply be $n+1$ . In the range of $y$ values from $1$ to $2015$ , the values of $n$ will be from $0$ to $2014$ , which is $2015$ numbers that will give a natural $y$ . The highest value of $x$ satisfying the conditions will be $2014 \cdot 2015 + 1$ and the lowest value of $x$ satisfying the conditions will be $0 \cdot 1 + 1$ . That gives $2014 \cdot 2015 + 1$ total possibilities for $x$ , and the probability of getting a natural value of $y$ is $\dfrac{2015}{4058211 }$ , and since that's in simplest form already, $a+b = \boxed{4060226}$ .

Let's try it in more algebraic way!!!

First, we have $y=\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-...}}}}$

$y=\sqrt{x+\sqrt{x-y}}$

$y^{2}-x=\sqrt{x-y}$

$y^{4}-2y^{2}x+x^{2}=x-y$

So, we will treat it as a polynomial in terms of x

$x^{2}-(2y^{2}+1)x+y^{4}+y=0$

$x=\dfrac{2y^{2}+1 \pm \sqrt{(2y^{2}+1)^{2}-4(y^{4}+y)}}{2}$

$x=\dfrac{2y^{2}+1 \pm \sqrt{4y^{4}+4y^{2}+1-4y^{4}-4y}}{2}=\dfrac{2y^{2}+1 \pm (2y-1)}{2}$ Since 2y-1 always be positive number.

Now, we have to choose only one root that correct (another root will exceed because the method of square). And after I've tried both two roots I found that the negative root (choose negative sign) is the answer.

Then we have

$x=y^{2}-y+1$

Since y vary from 1 to 2015, x will run from 1 to $2014 \times 2015+1$ but there are just 2015 value of x that can produce integer y.

Hence the probability is then $Prob=\dfrac{2015}{2014 \times 2015+1}$ and also this is the simplest fraction form.

$\dfrac{a}{b}=\dfrac{2015}{2014 \times 2015+1}$

$a+b=2015^{2}+1=4060225+1=\boxed{4060226}$

Let $\sqrt{x + \sqrt{x - \sqrt{x + ...}}} = A$ and $\sqrt{x - \sqrt{x + \sqrt{x - ...}}} = B$

From these equations, we can get $A = \sqrt{x + B}$ and $B = \sqrt{x - A}$

$A = \sqrt{x + B}$

$A^2 = x + B$

$x = A^2 - B$ -------------------------------------> $[equation 1]$

$B = \sqrt{x - A}$

$B^2 = x - A$

$x = B^2 + A$ -------------------------------------> $[equation 2]$

From equation 1 and equation 2, we will get

$A^2 - B = B^2 + A$

$A^2 - B^2 - A - B = 0$

$(A - B)(A + B) - (A + B) = 0$

$(A + B)(A - B - 1) = 0$

Since $A, B > 0$ , $A + B > 0$ . Therefore $A - B - 1 = 0$ or $B = A - 1$

Using equation 1,

$x = A^2 - B$

$x = A^2 - A + 1$

$0 = A^2 - A + (1 - x)$

But $A = y$

$0 = y^2 - y + (1 - x)$

This equation must lead to $[y + M][y - (M + 1)]$ for some natural number $M$ . The value of $y$ will be $M + 1$ .

$1 - x = -M(M + 1)$

$x = 1 + M(M + 1)$

$x = 1 + y(y - 1)$

Range of $y$ is $1 \leq y \leq 2015$ . Then the range of $x$ must be $1 \leq x \leq 2014 \times 2015 + 1$ . Therefore there are only $2015$ values of $y$ for $2014 \times 2015 + 1$ values of $x$ .

$\frac{a}{b} = \frac{2015}{2014 \times 2015 + 1}$

$a = 2015$ and $b = 2014 \times 2015 + 1$

$a + b = (2014)(2014 \times 2015 + 1) = 4060226$