mod use Show that the following equations have no integer solutions: (4) 25 + 12x2 +...
Show that the following equations have no integer solutions: (4) 25 + 12x2 + 24x + 1 = 0. (1) x4 + x + 1 = 0; (2) 12x + 5 = y2; (3) x? +6x +1 = 0;
Show that the following equations have no integer solutions: (4) 25 + 12x2 + 24x + 1 = 0. (1) x4 + x + 1 = 0; (2) 12x + 5 = y2; (3) x? +6x +1 = 0;
8.1. Show that the following equations have no integer solutions. a. 423-Туз-2003
8.1. Show that the following equations have no integer solutions. a. 423-Туз-2003
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.,...,
please answer question #5 and show steps
5. Solving Quadratics (mod p). Use #1 (a) above and the quadratic formula (mod p) to find a pair of solutions (if possible) for each of the following quadratic equations (1), 2r2 +3 -4- (mod 7) (ii), 3r2-2r +1 0 (mod 19) (ii). 3z2 2r -0 (mod 23) 1. Euler's Criterion. (a). Use Euler's Criterion to the Legendre symbols below: (iv). (10/23) (b). Assume a is a quadratic residue mod p, and assume...
2. Find 11644 mod 645 Use the following algorithm and show work! procedure modularExponentiation(b: integer, n = (ak-1ak-2...a1a0)2, m:positive integer) x:= 1 power := b mod m for i = 0 to k-1 If ai = 1 then x:= (x⋅power) mod m power := (power⋅power) mod m return x ( x equals bn mod m) Note: in this example m = 645, ai is the binary expansion of 644, b is 11.
1. Show that the number of solutions (x mod p, y mod p) to the equation x² + 1 = y2 mod p is p- p (6+1) k=0
5. (a) Show that 26 = 1 mod 9. (b) Let m be a positive integer, and let m = 6q+r where q and r are integers with 0 <r < 6. Use (a) and rules of exponents to show that 2" = 2 mod 9 (c) Use (b) to find an s in {0,1,...,8} with 21024 = s mod 9.
3) Show that the set of al numbers that are solutions of polynomial equations with integer coefficients is a countable set.
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.)