How to derive an proof for this infinite square root equation? Here is continuous square root, namely:
$\sqrt {1 + a \sqrt {1+b \sqrt {1+c\sqrt {1 +...}}}}$= any integer
Find $a,b,c,d,e,f,...$ in general
Uh, very interesting algebra pre-calculus problem, yet very challenging.
I know part of the answer but doesn't know how to start working on this problem.
The original problem is to prove $\sqrt {1 + 2 \sqrt {1+3 \sqrt {1+4\sqrt {1 +...}}}}$$=3$
However,i am curious on how to prove that we have finite or prove that we have infinite number of answer that satisfy the equation

Copying from Wikipedia in case of losing the data
Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':
$? = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \, $
This can be solved by noting a more general formulation:
: $? = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}}$
Setting this to F(x) and squaring both sides gives us:$
: $F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}$
Which can be simplified to:
: $F(x)^2 = ax+(n+a)^2 +xF(x+n) $
It can then be shown that:
: $F(x) = {x + n + a}$
So, setting ''a'' =0, ''n'' = 1, and ''x'' = 2:
: $3= \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}$
Ramanujan stated this radical in his lost notebook
$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$
