I am stuck on this problem for alcumus contest!

Here:

\(\sqrt{2-\sqrt{2-\sqrt{2-...}}}\)

Set it equal to x

\(\sqrt{2-\sqrt{2-\sqrt{2-...}}}=x\) equation 1

Square both sides

\(2-\sqrt{2-\sqrt{2-...}}=x^2\)equation 2

Because it goes on for infinity, it makes sense for us to susbtitute equation 1 into equation 2,

\(2-x=x^2\)

Now solve.

\(x^2+x-2=0\)

\(x=(-2,1)\)

So the answer is 1, as the answer can't be negative.

A trick to remember is this formula.

\(n-\sqrt{n-\sqrt{n-...}}=\frac{n}{2}\)

Does anyone know why it works? A proof?

Proof, please.

Here's what I  get  as a "proof"

Let  "a" be the integer under the root....using  CU's quadratic, we have that

a  -  x  = x^2

x^2  + x  -  a  =  0

x^2 + x  =   a              complete the square on  x

x^2 + x  + 1/4   =   a + 1/4

x^2  + x + 1/4  =  (4a + 1)  / 4

(x + 1/2)^2  =   (4a + 1) / 4         take the positive root

x  + 1/2  =    sqrt (4a + 1)  / 2

x =  sqrt  (4a + 1)  - 1

      _______________

                   2

Note   ...

If   a  = 2....then   this evaluates  to   1

If a  =  6.....then  this evaluates to  2

????