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location Canada
age 32
visits member for 2 years, 2 months
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I hold a diploma in Network Administration (along with many industry certifications) and am currently studying Computer Science and Pure Math in University.


Oct
28
comment If $\mathbb{Z}_m^*$ is cyclic, and $\mathbb{Z}_m^*=\langle\overline{g}\rangle$, is $\overline{g}$ a primitive root?
@ThomasAndrews I meant to imply when it is cyclic; as in, when $m=1,2,4$, or is of the form $p^{\alpha}$ or $2p^{\alpha}$ where $p$ is an odd prime.
Oct
28
asked If $\mathbb{Z}_m^*$ is cyclic, and $\mathbb{Z}_m^*=\langle\overline{g}\rangle$, is $\overline{g}$ a primitive root?
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@JyrkiLahtonen OK, then from the table alone, assuming I correct it to use multiplication instead of powers, can I find a generator from it? Or do I have to test each one individually with the method above?
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@JyrkiLahtonen Wouldn't that mean that every element in $\mathbb{Z}_9^*$ is a generator, save for $\overline{1}$?
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@JyrkiLahtonen But that seems contrary to what I was taught... but again I may have misunderstood
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@coffeemath notice the headers in my table. Are the headers wrong? I listed all the elements $n$ such that $(n,9)=1$
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
@coffeemath but I thought I had to take $2^4=16\equiv 7\mod{9}$
Oct
28
comment What are the generators of $\mathbb{Z}_9^*$?
Where are my errors? I've gone over this twice and I got the same thing each time.
Oct
28
asked What are the generators of $\mathbb{Z}_9^*$?
Oct
27
asked Find the natural numbers $n$ for which $\varphi(n)$ is not divisible by $4$.
Oct
27
revised Solve for n: $\varphi(2n)=\varphi(3n)$
deleted 5 characters in body
Oct
25
comment Solve for n: $\varphi(2n)=\varphi(3n)$
@lhf Updated the math
Oct
25
revised Solve for n: $\varphi(2n)=\varphi(3n)$
added 32 characters in body
Oct
25
comment Solve for n: $\varphi(2n)=\varphi(3n)$
@lhf I think you may be right, but I don't know how to show it... My math above used the assumption that if $(3,n)=3$, then $\varphi(3n)=\varphi(3^2n)=\varphi(3^2)\varphi(n)=3(3-1)\varphi(n)$... but now I see that I didn't factor the $3$ out of the $n$, so I should probably have $\varphi(n/3)$ instead...
Oct
25
answered Solve for n: $\varphi(2n)=\varphi(3n)$
Oct
24
comment Solve for n: $\varphi(2n)=\varphi(3n)$
All I was taught was that it's multiplicative.
Oct
24
asked Solve for n: $\varphi(2n)=\varphi(3n)$
Oct
23
accepted What is an “incongruent” solution?
Oct
23
comment What is an “incongruent” solution?
@labbhattacharjee Why is it not called "congruence" then, not "incongruence"?
Oct
23
comment What is an “incongruent” solution?
Does this mean that there are 11 different equivalence classes?