Nested Square Roots

Let $\displaystyle \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-.....}}}}=x$ .

Also, let $\displaystyle \sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+.....}}}}=y$ .

Then we see by observation that:-

$\displaystyle \large \sqrt{4+y}=x$

$\displaystyle \large \sqrt{4-x}=y$

Now whole square both equations and subtract to get:-

$\displaystyle x^2-y^2=x+y$ , which, in turn, means $\displaystyle x-y=1$ . Put $\displaystyle y=x-1$ in square of equation $\displaystyle 2$ , to get the final equation- $\displaystyle x^2-x-3=0$

Apply Sridharacharya's Rule (the quadratic formula) to get $\displaystyle x=\frac{1+\sqrt{13}}{2}$ . Note that $\displaystyle x$ can't be the other root $\displaystyle \frac{1-\sqrt{13}}{2}$ because $\displaystyle x$ is positive.

Finally, Comparing it with $\displaystyle \frac{a+\sqrt{b}}{c}$ , we get $\displaystyle a=1$ , $\displaystyle b=13$ , and $\displaystyle c=2$ , so-

$\displaystyle a+b+c=\boxed {16}$

Let $x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-...}}}}$ . Square both sides and substract $4$ :

$x^2-4=\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+...}}}}$

Square and substract $4$ again:

$(x^2-4)^2-4=-\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-...}}}}$

We need $x>2$ because the square roots of positive numbers only can be positive.

Note that the RHS is just $-x$ , so:

$x^4-8x^2+16-4=-x \Rightarrow x^4-8x^2+x+12=0$

Now, that equation factors as: $(x^2+x-4)(x^2-x-3)=0$

By solving each factor with the quadratic formula, we get four solutions:

$x_{1,2}=\dfrac{-1 \pm \sqrt{17}}{2}$

$x_{3,4}=\dfrac{1 \pm \sqrt{13}}{2}$

From these, the one that we want is $x=\dfrac{1 + \sqrt{13}}{2}$ , because is greater than $2$ . So, we have $a=1$ , $b=13$ , $c=2$ and $a+b+c=\boxed{16}$ .

I have seen something similar before. But a very straight forward way to do this is we say x =sqrt (4+Sqrt(4-x)) square both sides and subtract to get x^2-4=sqrt(4-x). Now we let this be a quadratic in 4. For completely psychological reasons we will let y=4, and proceed to square both sides and collect all terms as a quadrartic in y. What we get is the following y^2-(2x^2 +1)y+x^4+x=0. The fact that the answer is in the given form is a clue that it is factorable into two quadratics since it doesnt have any cubed or fourth roots. No we need two roots that add to 2x^2+1 and multiply to x^4+x. So we go about this by factoring x^4+x as x(x^3+1)=x(x+1)(x^2-x+1)=(x^2+x)(x^2-x+1). These work. So in fact for any y this thing will factor as (y-(x^2+x))(y-(x^2-x+1) at which point you plug in four and proceed with quadratic formula.

Let $\displaystyle \sqrt { 4+\sqrt { 4-\sqrt { 4+\sqrt { 4-... } } } } =S$

So $\displaystyle \sqrt { 4+\sqrt { 4-S } } =S$

Then \displaystyle 0={ S }^{ 4 }-8{S}^{4}+S+12

So $\displaystyle {0}=({S}^{2}-S-3)({S}^{2}+S+4)$

Now \displaystyle (S-\frac { 1 }{ 2 } )^{ 2 }-\frac { 13 }{ 4 } ={ S }^{ 2 }-S-3

Finally $\displaystyle S=\frac { 1 }{ 2 } +\sqrt { \frac { 13 }{ 4 } } =\frac { 1+\sqrt { 13 } }{ 2 }$ .