E.g., in $Z_9$, $1 = 2 * 5$.
Thanks!
Prove all "odd" elements of $Z_m$ (with m odd) can be written as $2 * x$
0
$\begingroup$
number-theory
elementary-number-theory
4 Answers
0
Since $m$ is odd, write $m = 2k - 1$. Now let $y$ be any integer, and let $\overline{y}$ be the corresponding element of $Z_m$.
Then $$y = 2(ky) - my \equiv 2(ky) \mod m.$$
Therefore if we let $x = ky$, we have $\overline{y} = 2\overline{x}$ in $Z_m$.
0
$k-m$ is even if m and k are odd, so $2\frac{k-m}{2}=k-m\equiv k (\mod m)$
0
Simple just show the stronger statement that all elements of $\Bbb Z_m$ can be written as $2\cdot x$. Since $\gcd(2,m)=1$ use the Euclidean algorithm to find $x,y$ so that $2x+my=1$. Then $2x\equiv 1\mod m$. Then let $a\in \Bbb Z_m$ and we have that $2xa=x2a=a$ so multiplication by $2$ is a bijection on $\Bbb Z_m$, i.e. it just permutes the elements. In particular $n\mapsto 2n$ is onto, so all elements, odd or otherwise, are of the form $2n$ for some $n\in\Bbb Z_m$.
0
You want to solve equation $2x = b$. Since $2$ is invertible in $\Bbb Z_m$ for odd $m$, you can always solve it: $x = 2^{-1}b$, or more explicitly, $2^{-1} = \frac{m+1}{2}$ in $\Bbb Z_m$, i.e. $x = \frac{m+1}{2}\cdot b$.