<s>[INST] 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
 [/INST] 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
$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$
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