FRACTURE

Yesterday, while scrolling through YouTube, I came across a video from a channel called blackpenredpen (bprp), where they broke down a very interesting problem that immediately caught my attention.

The problem was: if $\sqrt{a+\sqrt{a+\sqrt{a}}} = 128$, then $\lfloor a \rfloor = \ ?$

I found it quite inspiring, so I changed the value 128 to 69 and tried solving it myself using the same method demonstrated in the video.

Problem: $\quad \sqrt{a+\sqrt{a+\sqrt{a}}} = 69, \quad \lfloor a \rfloor = \ ?$

Solution: Because there are three nested radicals, isolating $a$ forces us to square the expression three separate times. This turns the problem into a massive high-degree polynomial equation.

$$ \begin{align*}\sqrt{a+\sqrt{a+\sqrt{a}}} &= 69 \\a+\sqrt{a+\sqrt{a}} &= 69^2 \\a+\sqrt{a+\sqrt{a}} &= 4761 \\\sqrt{a+\sqrt{a}} &= 4761 - a \\\left(\sqrt{a+\sqrt{a}}\right)^2 &= (4761 - a)^2 \\a+\sqrt{a} &= 4761^2 - 2(4761)(a) + a^2 \\a+\sqrt{a} &= 22667121 - 9522a + a^2 \\\sqrt{a} &= a^2 - 9522a - a + 22667121 \\\sqrt{a} &= a^2 - 9523a + 22667121 \\(\sqrt{a})^2 &= (a^2 - 9523a + 22667121)^2 \\a &= \left[(a^2) + (-9523a) + (22667121)\right]^2 \\ a &= (a^2)^2 + (-9523a)^2 + (22667121)^2 \\ &\quad + 2(a^2)(-9523a) + 2(a^2)(22667121) + 2(-9523a)(22667121) \\ a &= a^4 + 90687529a^2 + 513798374428641 \\ &\quad - 19046a^3 + 45334242a^2 - 431717986566a \\ a &= a^4 - 19046a^3 + (90687529 + 45334242)a^2 \\ &\quad - 431717986566a + 513798374428641 \\ a &= a^4 - 19046a^3 + 136021771a^2 \\ &\quad - 431717986566a + 513798374428641 \\ 0 &= a^4 - 19046a^3 + 136021771a^2 - (431717986566 + 1)a + 513798374428641 \end{align*} $$

$\therefore \ a^4 - 19046a^3 + 136021771a^2 - 431717986567a + 513798374428641 = 0$

This is completely solvable using the quartic formula, especially with strong computational tools. But realistically, no human brain is meant to carry that calculation out mentally—especially not mine.

So, let’s start from a similar equation that we can already solve, which is the infinite nested radical $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = 69$.

Since our problem has only three nested $a$ terms, and this equation has infinitely many $x$ terms; that means $x$ has to be less than $a$ for both expression to be equal to 69.

$$ \begin{align*}    \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} &= 69 \qquad [x < a] \\    x + \sqrt{x+\sqrt{x+\dots}} &= 69^2 \\[-0.4cm]    \underbrace{\qquad\qquad\qquad}_{\displaystyle 69} & \\    x + 69 &= 4761 \\    x &= 4761 - 69\end{align*} $$

$\therefore \ x = 4692 \implies \boxed{4692 < a} \qquad \left[ a \overset{?}{<} 4693 \right]$

$$ \begin{align*} \sqrt{4693 + \sqrt{4693}} &> \sqrt{4693 + \sqrt{68^2}} & \qquad \left| \, \begin{aligned} 68^2 &= 4624 \\ 68^2 &< 4693 \end{aligned} \right. \\[-.3cm] \sqrt{4693 + \sqrt{4693}} &> \sqrt{4693 + 68} \\ \sqrt{4693 + \sqrt{4693}} &> \sqrt{4761} \\ \sqrt{4693 + \sqrt{4693}} &> 69\\\end{align*} $$

$\therefore \ a < 4693 \implies 4692 < a < 4693 \implies \mathbf{\boxed{\lfloor a \rfloor = 4692}}$

The YouTube video ended just here. But I think I can go a step further by considering a general value $n$, instead of restricting myself to specific choices like 128 or 69.

Problem: $\quad \sqrt{a+\sqrt{a+\sqrt{a}}}=n,\quad \text{where } n \in \mathbb{Z}^+ \text{ and } n > 1.\quad \text{Find } \lfloor a \rfloor.$

Solution:

$$ \begin{align*}    \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} &= n \qquad [x < a] \\    x + \sqrt{x+\sqrt{x+\dots}} &= n^2 \\[-0.4cm]    \underbrace{\qquad\qquad\qquad}_{\displaystyle n} & \\    x + n &= n^2 \\    x &= n^2 - n\end{align*} $$

$\therefore \ x = n^2 - n \implies (n^2 - n) < a \qquad \left[ a \overset{?}{<} (n^2 - n + 1) \right]$

Since, $(n^2 - n + 1) > (n^2-2n + 1) = (n - 1)^2$

$$ \begin{align*}    \sqrt{(n^2 - n + 1) + \sqrt{(n^2 - n + 1)}} &> \sqrt{(n^2 - n + 1) + \sqrt{(n-1)^2}} \\    \sqrt{(n^2 - n + 1) + \sqrt{(n^2 - n + 1)}} &> \sqrt{(n^2 - n + 1) + (n-1)} \\    \sqrt{(n^2 - n + 1) + \sqrt{(n^2 - n + 1)}} &> \sqrt{n^2} \\    \sqrt{(n^2 - n + 1) + \sqrt{(n^2 - n + 1)}} &> n \\\end{align*} $$

$\therefore \ a < (n^2 - n + 1) \implies (n^2 - n) < a < (n^2 - n + 1)$

$\implies \boxed{\lfloor a \rfloor = (n^2 - n) = n(n - 1)}$

$\implies \boxed{\lceil a \rceil = (n^2 - n + 1) = n(n - 1) + 1}$