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 +...}}}}$$=3 However,i am curious on how to prove that we have finite or prove that we have infinite number of answer that satisfy the equation • are all of those variables integers? Oct 28, 2014 at 2:06 • @Asimov - List all possible answer in different forms (yep, it could be the integers.) Oct 28, 2014 at 2:07 • Surely a = 8, b = c = d = ... = 0 works fine. Oct 28, 2014 at 2:08 • Ok, so you want all sets of answers, not one set that falls under the integer constraint. Oct 28, 2014 at 2:09 • a=2,b=3,c=4,c=5,d=6... work also Oct 28, 2014 at 2:10 2 Answers This really should be a comment but it is too long. There are infinitely many periodic solution which returns 3. Let g(x) be the function x^2-1 and$$g^{\circ n}(x) = \underbrace{g(g(\ldots g(g(}_{n \text{ times}}x))\ldots))$$be the function obtained by composing g(x) with itself for n times. For any even n = 2k \ge 2, it is easy to check g^{\circ 2k}(3) is divisible by 3. One can verify$$(a,b,c,d\ldots) = (\; \underbrace{1, 1, \ldots, 1}_{(2k-1) \text{ times}}, g^{\circ 2k}(3)/3,\; \underbrace{\ldots}_{\text{ just repeat previous pattern}} )$$provides a periodic solution of length$n$. The first few examples, are •$n = 2$,$(a,b,\ldots) = (1,21, 1,21, \ldots )$. •$n = 4$,$(a,b,\ldots) = (1,1,1,5248341, 1,1,1,5248341, \ldots )$. •$n = 6$,$(a,b,\ldots) = (1,1,1,1,1, 20485753507127298001376466261, \ldots )$. 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 $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$

• What page on wiki? Oct 28, 2014 at 3:38