Answer to Question #169103 in Abstract Algebra for Jenna

Question

Let 𝑛 n be a positive integer that is at least 3 . Show that 𝑈(2^𝑛) has at least three elements of order 2. Prove using Bezout's theorem

Answer 1

U(m) is the group of positive integers j $\leq$ m such that gcd(j, m) = 1, under multiplication modulo m.

Since elements in u($2^n)$ are coprime to 2,

$U(2^n)= {{1, 3, 5...2^n-3, 2^n-1}}$ }The order

u($2^n)$ is $\varphi(2^n)=2^{n-1}, \\$

$2^{n-1}$ is of order 2, since

$(2^{n-1})^2=2^{2n}-2^{n+1} \equiv1mod2^n\\ Also\\ (2^{n-1}+1)^2=2^{2n-2}+2^{n}+1\equiv1mod2^n\\ More so\\ (2^{n-1}-1)^2=2^{n-2}-2^{n}+1\equiv1mod2^n\\ for \\ n\ge3$