Answer to Question #169103 in Abstract Algebra for Jenna

Question #169103

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


1
Expert's answer
2021-03-09T02:02:30-0500

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(2n)2^n) are coprime to 2,

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

u(2n)2^n) is Ο†(2n)=2nβˆ’1,\varphi(2^n)=2^{n-1}, \\

2nβˆ’12^{n-1} is of order 2, since

(2nβˆ’1)2=22nβˆ’2n+1≑1mod2nAlso(2nβˆ’1+1)2=22nβˆ’2+2n+1≑1mod2nMoreso(2nβˆ’1βˆ’1)2=2nβˆ’2βˆ’2n+1≑1mod2nfornβ‰₯3(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


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