# 2005 AIME II Problems/Problem 7

## Problem

Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}.$

## Solution

The expression for $x$ looks very suspicious. We multiply top and bottom by $(\sqrt[16]{5} -1)$.

$(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5} -1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)((\sqrt[16]{5})^2 - 1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5} - 1) =$ $= (\sqrt{5}+1)(\sqrt[4]{5}+1)((\sqrt[8]{5})^2-1) = (\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1) = (\sqrt{5}+1)((\sqrt[4]{5})^2-1) = (\sqrt5 + 1)(\sqrt5 - 1) = 4$. (This is an example of a telescoping expression. An alternative way to recognize the telescoping nature would be to write roots as fractional exponents, to write everything in terms of $\sqrt[16]5$, or to expand the denominator out entirely.)

Thus, we have immediately that $x= \frac{4(\sqrt[16]5 - 1)}{4} = \sqrt[16]5 - 1$ so $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125$