Find all solutions to the congruence x2+ x+ 1≡0 mod 91. (Hint:factor the modulus, use trial and error to find the solutions modulo the factors, and the CRT to combine the results into solutions to the original equations.)
Find all solutions to the congruence x2+ x+ 1≡0 mod 91. (Hint:factor the modulus, use trial and e...
39. Suppose that the polynomial congruence f(x)0 (mod 7) has two distinct so- 0 lutions, what are the possible number of solutions of the congruence f(x) (mod 49)? 39. Suppose that the polynomial congruence f(x)0 (mod 7) has two distinct so- 0 lutions, what are the possible number of solutions of the congruence f(x) (mod 49)?
15. Show that 716-1 (mod 17) and use that congruence to find the least non- negative residue of 7546 modulo 17
(b) Find the 2 elements x in Φ(289) such that x2 = 77 (mod 289) (Hint. First solve the congruence modulo 17.)
Solve the following system of equations and find all congruence class solutions if any exist; if no solution exists, explain why not: x24 (mod 35) 3.x 15 mod 21)
Problem 3. Use the Chinese Remainder Theorem to find all congruence classes that satisfy x2 = 1 mod 77.
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...
Please solve the above 4 questions. 1. Using the extended Euclidean Algorithm, find all solutions of the linear congruence 217x 133 (mod 329), where 0 x < 329 (Eg. if 5n, n 0,. ,6) 24 + 5n, п %3D 0, 1, . .., 6, type 24 + x< 11 2. Find all solutions of the congruence 7x = 5 (mod 11) where 0 (Eg. if 4,7 10, 13, type 4,7,10,13, none. or if there are no solutions, type I 3....
use of Theorelli 2.29. 3. Prove that x 14 + 12x2 = 0 (mod 13) has 13 solutions and so it is an identical congruence. -o(mod n) hasi solutions x = a, x = a.,...,
1. (Complex Multiplication) Let E : y x3 y23 to this congruence mod p. So for example, #E(Z3) = 3 because we have the solutions (0, 0), (1,0) and (2,0) and no more. - x. Then we can reduce E mod p to get mod p for various primes p. We write #E(Z») for the number of solutions This particular equation has some miraculous explore here patterns we (a) Make a chart that lists p, #E(Zp), and #E(Z) - p...
Find all solutions of the given congruence: x5 + 3x2 +1 = 0(125).