26 NUMBER THEORY FIND THE LEAST COMPLETE SOLUTION 15 rs x2 x

15. rs x2 x4 + x + 23 0 (mod 81) 17. z4+x +23 0 (mod 125) 18. x3 + 13 19. Show that if y / (mod p), then xk + pky + + 1 16. 0 (mod 25) 0 (mod 49) B 19. Show t + pty (mod pk+1 20. Prove by induction on n that if m and n are positive integers, then (u is claimed in the proof of Theorem 5.2. rove by induction on n that if m and n are positive integers, ther rnu)\" u\" + nun-imu (mod m2). In the neat siz problems, find the least complete solution. 21. x2-x-12 0 (mod 100) 22. 102.2-X3 9 (mod 405) 23. 2 12 (mod 1000) 2423 -50 0 (mod 275) 25\" x2-x + 7 0 (mod 63) (20 x2-x + 7 0 (mod 117) 27. Find all , 0

Solution

sol: the given equation x²-x+7 =o(mod 117)

now we have to find the least complete solution .prime factorization of 117 is 3*3*13 .

now 117 =3*3*13 / x2-x+7 implise that the polenomial is Reducible over F₃ and F₁₃ so there is a solution

under modulo 3 with x=2 and a solution under modulo 13 with x=3 or x=11.

science the moduls is composite ,so we solve the equation with mod(3*3*13).

therefore ,we have x=2(mod3)

x=3(mod13).

therefore by chinese remaindr theorem

x=107(mod39)

x=29(mod39).

so the complete solution is x= 29(mod39).