Primitive Root modulo n n=98

Asked Apr 03 '16 at 15:25

Active Apr 03 '16 at 15:29

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I would like to make sure I am understanding this correcting. My question is "Find primitive roots modulo $n$ using EGLF where $n=98$.

I have $\phi(98)=2(7^2)=98(1-\frac{1}{2})(1-\frac{1}{7})=42$ Since $(a,98)=1$ by EGLF $a^{\phi(n)} =a^{42}=1$. Therefore the primitive roots mod $n$ are $a^{42}$ and $a^{84}$. I'm saying $a^{84}$ because $42+42=84$ and is less then $98$. Does this look right?

edited Apr 03 '16 at 15:29

Ethan Hunt

1,033

asked Apr 03 '16 at 15:25

Natasha Epperson

2

What does EGLF mean? – Ethan Hunt Apr 03 '16 at 15:29
HAHA my bad, Euler's generalization of little Fermot – Natasha Epperson Apr 03 '16 at 15:32
Let a∈Z and (a,n)=1 then a^(ϕ(n))=1 (mod n), moreover ord_n a | ϕ(n). – Natasha Epperson Apr 03 '16 at 15:33
EGLF is usually called Euler's theorem. – Wojowu Apr 03 '16 at 15:49
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See http://math.stackexchange.com/questions/754187/prove-that-3-is-a-primitive-root-of-7k-for-all-k-ge-1 – lab bhattacharjee Apr 03 '16 at 15:50
I don't quite get it here: what is $;a;$, anyway? Any number coprime with $;98;$ ? It can't be, since $;a=1;$ trivially won't work, so what is it? – DonAntonio Apr 03 '16 at 16:13
yah (a,98)=1 what what EGLF says. a would be any number coprime with 98 by EGLF – Natasha Epperson Apr 03 '16 at 17:24

Primitive Root modulo n n=98

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