Use Fermat's little theorem to compute x^y^z mod p (where p is prime)?

Question

I'm new to Fermat's theorem (a, and I am familiar with how to use it in basic cases with relatively large numers (i.e. 2^345 mod 31). I was given the question to find 4^(2^2006), and found myself stuck with wrong answers (when comparing to the result on WolfRamAlpha of 8. How would I perform a calculation in the form of x^y^z mod m using Fermat's little theorem? Anything helps. Cheers!

Edit: What I have tried was as follows:

4^2²⁰⁰⁶ --> exponent: 2²⁰⁰⁶

exponent's exponent: 2006

Since I want mod 31, I attempted 2006 mod 31 = 22, then one level down...

2²² mod 31 = 4, then again, the last level down...

4⁴ = 256 mod 31 = 8.

While 8 is the correct answer, I do not believe this is the correct method to get the solution.

Could you briefly explain specifically what you tried? That would help us to compose better answers $\ddot\smile$ Also, here's the MathJax tutorial! — gen-ℤ ready to perish, Oct 16 '17 at 15:52
You must specify what you want.Example: $2^{345}≡1 mod 31$ . You must ask: $ 2^{2^{2006}}≡ r mod x$. tell us what is the value of x. — sirous, Oct 16 '17 at 16:38
@J.R. In general, one wouldn’t use Fermat’s little theorem twice. One of the two applications will typically be Euler’s theorem (or else multiple uses of Fermat followed by CRT). — Erick Wong, Oct 16 '17 at 22:00
@J.R. It’s clear to me as well, but much less so to someone who is new to Fermat’s little theorem and prone to making that type of mistake (of using the same prime modulus at each level). — Erick Wong, Oct 17 '17 at 00:53
@ErickWong You are right. Thanks for clarifying it for the OP! — J.R., Oct 17 '17 at 01:13

score 0 · Answer 1 · 2017-10-18T01:10:11.043

i suck at latex sry if someone could format this for me that would be great i believe this is a sol. not using f.l.t. as far as i can tell... i wonder if there is a faster way of doing it? it was like i just looked for the 1st power of 4 that =1 then divided out that power over and over till it 0ed out. what what was left was the answer.

4^15 mod 31 =1 4^(2^4) (mod 31)≡ 4^16 mod 31

≡4^15 * 4 mod 31

≡ (4^15 mod 31 * 4 mod 31)
≡ (1 * 4 mod 31) = 4

4^(2^2006 ) (mod 31)≡ 4^(2^2004* 2^2 ) (mod 31)

≡(4^(2^4 ))^(2^2000*2^2 ) (mod 31)

≡ 4^(16)^(2^2000 2^2 ) (mod 31)

≡ (4^15 * 4 )^(2^2000* 2^2 ) mod 31 ≡ (4^15 (mod 31) * 4 (mod 31) )^(2^2000* 2^2 )

≡ (1 * 4 (mod 31) )^(2^2000* 2^2 )

≡ 4^(2^2000* 2^2 ) mod 31

≡4^(2^4 *2^1998 ) (mod 31) ) ≡(4^15 * 4 (mod 31) )^(2^1998 ) ≡(4^(2^1998 ))(mod 31)

≡4^(2^4 *2^1994) (mod 31) ) ≡(4^15 * 4^1 (mod 31) )^(2^1994 )

≡4^(2^1994 )(mod 31)≡ 4^((2^1992)*2^2)

Continue in this way . . . ≡ 4^(2^0*2^2) ≡(4^(2^2 )) (mod 31) ≡(4^4) (mod 31) ≡8 Hence 4^(2^2006 ) (mod 31)≡8

Use Fermat's little theorem to compute x^y^z mod p (where p is prime)?

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