# x^7 = 12 mod 29 (primitive roots)

• April 23rd 2010, 02:56 PM
santiagos11
x^7 = 12 mod 29 (primitive roots)
I need to solve the following problem:
"=" means is congruent to.
x^7 = 12 mod 29

This is what I have:
I am using q=2 as the primitive root mod 29.
7*I(x)=I(12) mod 28
7*I(x)=7 mod 28
I(x) = 1 mod 4

What do I do next?
• April 23rd 2010, 06:53 PM
santiagos11
never mind, I got it. i will post the solution later
• April 23rd 2010, 06:57 PM
chiph588@
Well $(x,29)=1$, so we know $x\equiv2^y\bmod{29}$.

Therefore $\text{ind}_2(2^y) = y\equiv1\bmod{4}$

Our solutions are thus $x=\{2^1,2^5,2^9,2^{13},2^{17},2^{21},2^{25}\}$.

Note that these are the only solutions since $2$ is a primitive root, so $\{2^0,\ldots,2^{28}\}$ generate all possibe solutions.