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?

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- Apr 23rd 2010, 02:56 PMsantiagos11x^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? - Apr 23rd 2010, 06:53 PMsantiagos11
never mind, I got it. i will post the solution later

- Apr 23rd 2010, 06:57 PMchiph588@
Well $\displaystyle (x,29)=1 $, so we know $\displaystyle x\equiv2^y\bmod{29} $.

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

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

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