For how many integers $n$ between $ 1$ and $2021$ does the infinite nested expression $$\sqrt{n + \sqrt{n +\sqrt{n + \sqrt{...}}}}$$give a rational number?
Problem
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Tags: algebra, Nested Radicals, radical
fuzimiao2013
16.11.2021 00:38
I assume that all of them converge, not sure how to prove convergence, but just putting this notice here just in case jasperE3 sees this
Let $k = \sqrt{n + \sqrt{n +\sqrt{n + \sqrt{...}}}}$. Thus, $k = \sqrt{n+k} \implies k^2-k-n=0$, and so $n = (a)(a+1)$ for some positive integer $n$. Thus, $n$ can $1 \cdot 2$, $2 \cdot 3$, ... $44 \cdot 45$, meaning there are $44$ $n$.
greenturtle3141
16.11.2021 00:56
It converges for all positive integers $n$. All you need to show this is to prove two things: 1. As you add more radicals, the expression increases (not hard) 2. All the numbers $\sqrt{n}$, $\sqrt{n+\sqrt{n}}$, etc. are less than some common number. Here it can choose $n$ as the upper bound and prove it by induction.
jasperE3
16.11.2021 01:30
fuzimiao2013 wrote: jasperE3 sees this
$k^2-k-n=0$, and so $n = (a)(a+1)$
Hi, can you please clarify this part of the solution?
fuzimiao2013
16.11.2021 01:37
jasperE3 wrote: fuzimiao2013 wrote: jasperE3 sees this
$k^2-k-n=0$, and so $n = (a)(a+1)$
Hi, can you please clarify this part of the solution? The only way for the linear coefficient to be -1 while the constant term is negative is when you have -(a+1) + a = -1
oralayhan
17.11.2021 15:32
Official solution; Denote the expression by $x$. Then $x^2 = n+x$ so $$x^2-x-n = 0$$and $x = (1+\sqrt{1+4n})/2$. This is rational if $1+4n = m^2$ or $$4n = (m-1)(m+1)$$. Hence $m$ can be any odd number with $5<=m^2=<8085$, i.e., $$3<=m<=89$$(note $902 = 8100 > 8085$). There are $44$ such values of $m$, corresponding to $44$ values of $n$.