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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

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  • $\begingroup$ are all of those variables integers? $\endgroup$
    – Asimov
    Oct 28, 2014 at 2:06
  • $\begingroup$ @Asimov - List all possible answer in different forms (yep, it could be the integers.) $\endgroup$
    – Victor
    Oct 28, 2014 at 2:07
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    $\begingroup$ Surely $a = 8, b = c = d = ... = 0$ works fine. $\endgroup$
    – John Hughes
    Oct 28, 2014 at 2:08
  • $\begingroup$ Ok, so you want all sets of answers, not one set that falls under the integer constraint. $\endgroup$
    – Asimov
    Oct 28, 2014 at 2:09
  • $\begingroup$ a=2,b=3,c=4,c=5,d=6... work also $\endgroup$
    – Victor
    Oct 28, 2014 at 2:10

2 Answers 2

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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

  • $n = 2$, $(a,b,\ldots) = (1,21, 1,21, \ldots )$.
  • $n = 4$, $(a,b,\ldots) = (1,1,1,5248341, 1,1,1,5248341, \ldots )$.
  • $n = 6$, $(a,b,\ldots) = (1,1,1,1,1, 20485753507127298001376466261, \ldots )$.
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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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  • $\begingroup$ What page on wiki? $\endgroup$
    – apnorton
    Oct 28, 2014 at 3:38

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