Use group theory to calculate $9^{74} \pmod {13}$.

Question

By Fermat's theorem (or more generally by Euler's totient function theorem), for $p$ prime, any nonzero element $a$ of $Z/pZ$ satisfies:

$$a^{p-1}\equiv1 \mod p$$

Hence, in this case, we have $$9^{74}=(9^{12})^69^2$$

Applying the theorem:

$$9^{12}\equiv1 \mod 13$$

So

$$9^{74}\equiv9^2 \mod 13$$

And now, if I'm doing the math correctly,

$$81=(13)(6)+3$$

So

$$81\equiv3 \mod 13$$

Is this correct?

This is a special case of modular order reduction as described in the linked dupe. If you know some group theory you can reformulate that in more abstract language (I avoided that to keep it accessible to more readers). See the linked questions there for many more examples. — Bill Dubuque, Mar 31 '20 at 01:55

score 1 · Answer 1 · 2020-03-31T00:10:02.590

1

$\phi(13)=12$. Fermat's little theorem gives $9^{12}\cong1\pmod{13}$. Hence $9^{74}\cong9^2\cong3\pmod{13}$.

That is, you got it.

edited Mar 31 '20 at 00:10

answered Mar 31 '20 at 00:03

1 Answers1