Question

Exercise 4.9.29: Solve the following systems of congruences, or state that there is no solution. Be sure to state if there are multiple solutions.

a. {6 = 13a + b(mod 26), 13 = 4a + b(mod 26)

b. {14 = 17a + b(mod 26), 8 = 7a + b(mod 26)

c. {1 = 15a + b(mod 26), 10 = 9a + b(mod 26)

Answer 1

Solution:-

Given that

(a)

13a + b = 6 (mod 26)

4a + b = 13 (mod 26) ..........(1)

(13a + b) - (4a + b) = 6 - 13 (mod 26)

9a = -7 + 26 (mod 26)

= 57

= 5 (mod 26)

now, 4.5 + b = 13 (mod 26)

b = 13 - 20 + 26 (mod 26)

b = 19 (mod 26)

Hence

a = 5 + 26R,  

b = 19 + 26m,

are the solution of (1)

(b)

17a + b = 14 (mod 26)

7a + b = 8 (mod 26) .......................(2)

(17a + b) - (7a + b) = 14 - 8 (mod 26)

10a = 6 (mod 26)

= 3 . 21

= 11 (mod 26)

now, 7 . 11 + b = 8 (mod 26)

b = 8 - 77

= 8 - 77 + 3.26 (mod 26)

= 8 + 1 = 9 (mod 26)

Hence a = 11 + 26 R,

b = 9 + 26m,   

are the solution of (2)

(c)

15a + b = 1 (mod 26)

9a + b = 10 (mod 26) .................(3)

(15a + b) - (9a + b) = 1 - 10 (mod 26)

6a = -9 = 17 (mod 26)

As gcd (6, 26) = 2 1

Hence 6 has no inverse in

Hence 6a = 17 (mod 26) has no solution

Hence (3) has no solution