When Towers Meet Radicals

Let $x=2+\cfrac{a}{2+\cfrac{a}{2+\cfrac{a}{2+...}}}$

$x=2+\cfrac{a}{x}$

$x^{2}=2x+a$

$x^{2}-2x=a$

$x^{2}-2x+1=a+1$

$(x-1)^{2}=a+1$

$x-1=\pm\sqrt{a+1}$

$x=\pm\sqrt{a+1}+1$

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Let $y=\sqrt{a\sqrt{a\sqrt{a\sqrt{a...}}}}$

$y=\sqrt{a}\cdot\sqrt{\sqrt{a}}\cdot\sqrt{\sqrt{\sqrt{a}}}...$

$y=a^{\cfrac{1}{2}}\cdot a^{\cfrac{1}{4}}\cdot a^{\cfrac{1}{8}}\cdot ...$

$y=a^{\cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+...}$

$y=a^\cfrac{\cfrac{1}{2}}{1-\cfrac{1}{2}}$

$y=a$

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From $y$ , we know that $a\geq0$

$x>y\geq0$

$x>a\geq0$

$x>0$

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When $x=+\sqrt{a+1}+1$ ,

$+\sqrt{a+1}+1>0$ for $a\geq0\geq-1$

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When $x=-\sqrt{a+1}+1$ ,

$-\sqrt{a+1}+1>0$

$\sqrt{a+1}<1$

$a+1<1$

$a<0$ , which is contadictory to our claim that $a\geq0$ .

So, $x\neq-\sqrt{a+1}+1$

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Hence, $x=+\sqrt{a+1}+1$

$\sqrt{a+1}+1>a$

$\sqrt{a+1}>a-1$

There is a concern with such manipulation: If $p>q$ , $p>0, q<0$ , but $|p|<|q|$ , then $p^{2}<q^{2}$ ,

However, when $a>0$ , such $|\sqrt{a+1}|>|a-1|$

Therefore, we could go ahead with the manipulation with further squaring of both sides, giving

$(\sqrt{a+1})^{2}>(a-1)^{2}$

$a+1>a^{2}-2a+1$

$a^{2}-3a<0$

$a(a-3)<0$

Previously, we have shown that $a\geq0$ ,

So, it follows that $a<3$

The largest integer value of $a$ which satisfies the inequality is $a=2$