How do I use Fermat's Little Theorem to solve the following:

Question

$x^{13}$ = 2 mod 23. I know how to use the Fermat's theorem, but in this case, we don't know the value of "a". So I am unsure of how to even start?

Here all I know is that p = 23 and thus $a^{22}$ = 1 mod 23

What is the problem to be solved here? Find $x$? – bjorn93 Nov 07 '19 at 23:18 — bjorn93, Nov 07 '19 at 23:18
Yeah it just says to solve the equation – Nov 07 '19 at 23:34 — , Nov 07 '19 at 23:34

score 1 · Answer 1 · answered Nov 07 '19 at 23:21

1

Since $13\times17=221\equiv1\bmod22, $

$x^{13}\equiv2\bmod23\implies x^{13\times17}\equiv x\equiv 2^{17}\bmod 23.$

answered Nov 07 '19 at 23:21

J. W. Tanner

60,406

Note: $x^{13\times17}\equiv x$ follows from Fermat's Little Theorem, which says $x^{22}\equiv1\bmod23$; $2^{11}\equiv1\bmod23$ so $x\equiv2^6=64\bmod 23$ – J. W. Tanner Nov 07 '19 at 23:31
So this means $64^{13}$ divided by 23 will give me a remainder of 2? – Nov 07 '19 at 23:58
@SanchitKumar: that is correct – J. W. Tanner Nov 08 '19 at 00:02
Ok so is there a formal way you came up with that 17? – Nov 08 '19 at 00:10
I inverted $13$ modulo $22$ – J. W. Tanner Nov 08 '19 at 00:12
Oh so just the multiplicative inverse of 13 modulo 22 will result in 17 and then you multiply both sides by $x^{17}$? – Nov 08 '19 at 00:15
I took both sides to the $17^{th}$ power to get $x$ – J. W. Tanner Nov 08 '19 at 00:18
So by raising the left side to the 17th power we essentially eliminated the 13th power because 17 is the modulo inverse and we just got x??? – Nov 08 '19 at 00:36
I don't understand how you got 2$^{11}$ power – Nov 08 '19 at 00:39
Click on the link for that above (in my first comment) – J. W. Tanner Nov 08 '19 at 00:44
So how are we supposed to come up with that if we didn't know that before? – Nov 08 '19 at 00:53
$2^{11}\equiv1\bmod 23$ was not essential to the answer, just a short-cut to calculating the answer $2^{17}\bmod23$ – J. W. Tanner Nov 08 '19 at 00:59
So without the shortcut, how would we apply Fermat's little theorem to $(2^{17})$ – Nov 08 '19 at 01:06
Fermat’s little theorem was used to prove that $(x^{13})^{17}\equiv x$. There are many posts on this web site explaining how to compute something like $2^{17}\bmod23$ – J. W. Tanner Nov 08 '19 at 01:14
I think the answer should be 18 and not 64 the answer to $2^{17}$ is 18 not 64 – Nov 08 '19 at 01:30
mod $23$, $18$ and $64$ are the same (I had left reducing $64$ mod $23$ for you to do; note $64=23\times3-5$ and $18=23\times1-5)$ – J. W. Tanner Nov 08 '19 at 01:33
So what would be the steps on how to solve questions like these? Like for a general case, can you tell me what I would have to do step by step – Nov 08 '19 at 07:08
if $x^a\equiv c\bmod p$ and $ab\equiv1\bmod p-1$ then $x\equiv c^b\pmod p$ – J. W. Tanner Nov 08 '19 at 14:25
Oh okay. So what are these types of questions called? $x^a$ = c mod p? So I can learn more about them and also is there another example I may try like this? – Nov 08 '19 at 15:06
Here is a related question – J. W. Tanner Nov 08 '19 at 15:17

How do I use Fermat's Little Theorem to solve the following:

1 Answers1