Question

How many classes of solutions are there for each of the following congruences?

(a) x² - 1 = 0 mod (168)

(b) x² + 1 = 0 mod (70)

(c) x² + x + 1 = 0 mod (91)

(d) x³ + 1 = 0 mod (140)

Please note to show how you got the solutions as well as finding out how many classes of solutions there are for each congruence.  

Please explain every step so I can understand how tonot only the solutions but the classes of solutions.   

Answer 1

23 II n = ti 168 = 34 7x8 x²_1=0 (mod 3) han sols : or x = 1, 2 (mod 3) x-1=0 brood 7) hon sochs: x = lg6 modt) x-1=0 forod 8) han sols :6=113 on x= 1, 3, 5,7 (mod 8) 3, 7, 8 are muntually coprime, by the Chinese Remainder Theorem, The system of equation aza, (mod 3)? n = a (mod 7) has unique soen mord 3*7*8=168 and x = a (mod 8) fax= 1 or 2) 2 ways (an=1006) can be chosen in an 1 2 ways an 4 ways (93= 1,3,5 or 7) Total of solus وز 2x2x4 =16 70 = 2 6 x+1=0 food 70) 2 x 5 x 7 x +1 = o(mord 2) 2=1(moda) 241 o(shod 5) > x=+2,0;e, x=2, *3+2,0;:9, X=2,3(mods) xt = 0 mod a) → No soledion onodulo 7 x 4 =0 (mod 70) has opard the price = (2-1) (arat) a no soch © 91 - 7X13 x²1 food 7) 242, 4 (mod 7) = xxx+1=0 (mod 7) => x = 294 (mod 7) 2²31 (mod 13) => x=1, 3, 9 (mod 13) Xtati o (modt) has gots x= 3,{mod 15) by CRT, xenni (mod 81) has 2 x 2 = 4 solution. ²1 Eolmed (40) has 1x1x3=3 solution Q 1403 - 2x5x7 By CRT, & 120/mod 2) = x= 1 (mod 2) x² + 1 =oland 5) x = 2 (mod 5) x²*1 = o(mod 7) = x= 35,6 (mod)