Question

4 [20 points] Given that 491 is a prime and 26 is primitive modulo 491, use the Pohlig- Hellman algorithm to solve 26 192 (mod 491). Be sure to show your work. You may need the following data. 2624 490 (mod 491) 26 101 (mod 491) 2610 414 (mod 491) 414 223 (mod 491) 4141 51 (mod 491) 1922451 (mod 491) 1928 381 (mod 491) 19210 3 (mod 491) 223 153 (mod 491)

Answer 1

we see 491 - 1 = 490 = 2 x 5 x 7².

we determine the numbers

x_{​​​​​​}​​​​​1 = x mod 2

x_{​​​​​}​​​​​​2 = x mod 5

x_{​​​​​​}​​​​​​3 = x mod 7²

Determination of x_{​​​​​​}​​​​​1​​​​​ :

x1 = c0 mod 2 where c0 = 0 or 1

Now 192^(491-1)/2 = 192²⁴⁵ = 1 mod 491 = 26^245c0 mod 491 = 490^c0 mod 491.

So c0 = 0. That is x = 0 mod 2.

Determination of x2​​​​​ :

x2 = c0 mod 5 where c0 = 0,1,2,3 or 4

Now 192^(491-1)/5=192⁹⁸ = 381 mod 491 = 26^98c0mod 491 = 101^c0 mod 491.

So c0 = 2. That is x = 2 mod 5.

Determination of x3 :

x3 = c0 + c1(7) where c0,c1 = 0,1,2,3,4,5 or 6.

Now 192^(491-1)/7 = 192⁷⁰ = 3⁷ = 223 mod 491 = 26^70c0 mod 491 = 414^7c0 mod 491 = 223^c0 mod 491.

So c0 = 1.

Now 26^1+7c1 = 192 mod 491

=> 26^10+70c1 = 192¹⁰ mod 491 = 3 mod 491

=> 414 x 223^c1 = 3 mod 491

=> 223^c1 = 3(414)^-1 mod 491 = 3 x 51 = 153 mod 491

So c1 = 5 . That is x = 1 + 5x7 = 36 mod 49.

now,

x = 0 mod 2

x = 2 mod 5

x = 36 mod 49

M_{​​​​​​1} = 490/2 = 245 . 245 ^-1 = 1 mod 2.

M_{​​​​​​2} = 490/5 = 98 . 98 ^-1 = 2 mod 5

M_{​​​​​​3} = 490/49 = 10 . 10 ^-1 = 5 mod 49.

so by Chinese remainder theorem,

x = 0x245x1 + 2x98x2 + 36x10x5 mod 491

= 228 mod 491.

Hence x = 228.

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