Question

Problem 3. Use the Chinese Remainder Theorem to find all congruence classes that satisfy x2 = 1 mod 77.

Answer 1

5012 x² = 1 (mod 77) Then - x² = (mod 7) and x² = 1 (mod 11) Now, x² = 1 (mod 7) has only solt x = /(mod 7) similarly, x²= 1 (mod 11) has soft as x=+1 (mad 11) Now , use chinease remainder Them to find the 56/ 4 pair of egths (a) x = 1 (mod 7) (6) n = -1 (mod 7) (e) *= 1 (moda) (mod 11) 1 fmod11) me 3 -1 (mod 11) and (d) x= -1 (modt) x= -1 (mod 11) Then the soft if M = 77 > m, = 7, M = 11 to x = ai (mod mi) is given by- u= a, b, M + az b. M (mod M) m M2 where, preco M bi mi = 1 (mod mi) i=1,2. M be a ron (a) bi = 1 (mod m.) → b;11 = 1 (mod 7) 6:11 = 1 (mod 7) = b = 2 = 1 (mod wa) bz: 7 = 1 (mod 11) b₂ = 8 Therefore, x= (1+2× 11 + (1x2 x 1 + 1*8* 7)(mod 77) 78 (mod 77) x= 1 mod 77 25

& 6 (modt) x= 1 (mod 11) (6 25 x= (6x2x11 + 1x8*7) (mod 77) 188 (mod 77) → 2 = x= 34 mod 77 (c) 2 = 1 (mod) & x= 10 (mod 11) x = (1x2x11 + 10x887) (mod 77) x = 582 (mod 77) x= 43 (mod 77) 1 (d) 23 & 6 (moda) as 10 (mod 7) x= (6x2x11 & 10x8x7) (mod 77) 692 mod 77) x = 76 (mod 77) a ने M =