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Subject: "Discrete logarithm problem"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #662
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BenVitale
Member since Dec-20-07
Dec-20-07, 08:08 AM (EST)
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"Discrete logarithm problem"
 
   They say that the Discrete logarithm problem is supposed to be very difficult to solve. Here's an example:

Given the primitive element 2, the prime number 331, and the number 141, solve for the number x below:

2^x = 141 mod (331)

Can you find the value for x?

I found that 2 is not primitive mod 331, as 2^30 = 1 mod 331.

So it'seems that there is no solution.

This is what i saw. Does anyone see another solution?


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mr_homm
Member since May-22-05
Dec-21-07, 01:35 PM (EST)
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1. "RE: Discrete logarithm problem"
In response to message #0
 
   I get the same result. There are always some numbers that occur as powers of 2, even though 2 is not primitive. In this case, since the nonzero integers mod 331 are a group of order 330 under multiplication, the possible subgroups have orders 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165, and 330. As you observed, the powers of 2 generate a subgroup of order 30, so there is a 1 in 11 chance that a randomly chosen positive integer mod 331 would occur as a power of 2. As it happens, 141 is not there. Thus there is no solution to this problem.


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Pierre Charland
Member since Dec-22-05
Jan-20-08, 00:16 AM (EST)
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2. "RE: Discrete logarithm problem"
In response to message #0
 
   You're right, direct computation gives (mod 331):
2^0 = 1 = 1
2^1 = 2 = 2
2^2 = 4 = 4
2^3 = 8 = 8
2^4 = 16 = 16
2^5 = 32 = 32
2^6 = 64 = 64
2^7 = 128 = 128
2^8 = 256 = 256
2^9 = 512 = 181
2^10 = 1024 = 31
2^11 = 2048 = 62
2^12 = 4096 = 124
2^13 = 8192 = 248
2^14 = 16384 = 165
2^15 = 32768 = 330
2^16 = 65536 = 329
2^17 = 131072 = 327
2^18 = 262144 = 323
2^19 = 524288 = 315
2^20 = 1048576 = 299
2^21 = 2097152 = 267
2^22 = 4194304 = 203
2^23 = 8388608 = 75
2^24 = 16777216 = 150
2^25 = 33554432 = 300
2^26 = 67108864 = 269
2^27 = 134217728 = 207
2^28 = 268435456 = 83
2^29 = 536870912 = 166
2^30 = 1073741824 = 1
... (we have a 30-cycle)
if p is prime, phi(p)=(p-1),
phi(331)=330, 330/30=11,
so 2 has index 11 modulo 331,
so 2 is not primitive modulo 331.
(where phi is the Euler phi function)

AlphaChapMtl


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