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Fermats n Eulers theorem

Posted on 2004-09-28
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I know what Fermat's and Eulers theorem are.

this is a question that i have been given..

Use Fermat or Euler's theorem (whichever is applicable) to compute

a) 3 ^ 201 mod 11
b) 3^{2161} mod 1147..

how am i supposed to use Fermat's or Euler's thoerem for this?

Fermat's little theorem said that a^p-1 = 1 mod p.

any ideas?

thanks,
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Question by:younoeme
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by:d-glitch
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 3^201 mod 11   =   [(3^10)^20] * (3^1)  mod 11

                         =   [ 1^20 mod 11 ] * [3 mod 11]

                         =   3 mod 11

               
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by:d-glitch
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3^2161 mod 1147   =   3^1146  * 3^1015  mod 1147

                             =      1         * 3^512 * 3^256 * 3^128 * 3^64 * 3^32 * 3^16 * 3^4 * 3^2 * 3^1    mod 1147

                             =   [ more math to do                                 ]   *    958   *  826   *  81 *    9   *    3


You can use fermat's theorom to reduce the exponent to less than p-1

Whatever is left you have to the old fashioned way. But you can break it up into exponents that are powers of two, you can bootstrap your way up and reduce the math.

I did the first few.    The next step is:         3^64 mod 1147  =  3^32 ^ 3^32 mod 1147  =  958 * 958 mod 1147
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by:d-glitch
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I mad a mistake in my last post:

3^1   mod 1147    =     3
3^2    mod 1147   =     9
3^4    mod 1147   =    81
3^8    mod 1147   =  826  ==>  You don't use this term in the answer, but you still have to calculate it to get the next term in the series.
3^16   mod 1147  =  958
3^32   mod 1147  =  958^2 mod 1147
3^64   mod 1147
3^128  mod 1147
3^256  mod 1147
3^512  mod 1147


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by:d-glitch
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3^{2161} mod 1147..

1147 = 31 * 37   It isn't prime so you can't use Fermat's Little Theorem.
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by:d-glitch
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You need to use Euler's Totient Function:        http://mathworld.wolfram.com/EulersTotientTheorem.html
                                                                  http://mathworld.wolfram.com/TotientFunction.html


 phi(1147) = 1147 * (30/31) * (36/37)  = 1080   ==>  Euler's Totient Function

3^1080 mod 1147 = 1                                      ==> Euler's Totient Theorem


3^2161 mod 1147   =   (3^1080) * (3^1080) * 3^1  mod 1147
                             =        1        *       1       *  3     mod 1147
                             =    3
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by:rjkimble
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Verification from Maxima:

(C1) mod(3^201,11);
(D1)                                3
(C2) mod(3^2161,1147);
(D2)                                3
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d-glitch earned 100 total points
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In summary:

   a^n  mod b ==> mod( a^n, b)


If the base b is prime, you can use Fermat's Little Theorem to reduce the exponent:
 
                                        mod( a^n, b) = 1 reduces to mod( a^m, b) where m = mod(n, b-1)


If the base b is composite, you can use Euler's Totient Theorem to reduce the exponent:
 
                                        mod( a^n, b) = 1 reduces to mod( a^m, b) where m = mod(n, phi(b))

If m is small you may be able to find the answer by inspection.

If m is large, you can use the binary decomposition, as I suggested in my otherwise irrelevent post.
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by:rjkimble
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Very nicely stated.
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by:d-glitch
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Thanks rjk.

It would have been even nicer if I had left out the =1's

                                   mod( a^n, b)  reduces to ...

Another senseless cut and paste tragedy.
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by:rjkimble
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Good point. My bleary eyes glossed over that. Still, the way you laid things out is quite nice.

.... Bob
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