we see 491 - 1 = 490 = 2 x 5 x 72.
we determine the numbers
x1 = x mod 2
x2 = x mod 5
x3 = x mod 72
Determination of x1 :
x1 = c0 mod 2 where c0 = 0 or 1
Now 192(491-1)/2 = 192245 = 1 mod 491 = 26245c0 mod 491 = 490c0 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 = 2698c0mod 491 = 101c0 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 = 2670c0 mod 491 = 4147c0 mod 491 = 223c0 mod 491.
So c0 = 1.
Now 261+7c1 = 192 mod 491
=> 2610+70c1 = 19210 mod 491 = 3 mod 491
=> 414 x 223c1 = 3 mod 491
=> 223c1 = 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
M1 = 490/2 = 245 . 245 -1 = 1 mod 2.
M2 = 490/5 = 98 . 98 -1 = 2 mod 5
M3 = 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.
4 [20 points] Given that 491 is a prime and 26 is primitive modulo 491, use the Pohlig- Hellman a...