Question

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)

Answer 1

To solve the quadratic congruence a (mod p), one way is to compute the entire table of values for 2 modulo p. For very small prime such as the simple example above, this approach is workable. For large primes, this requires a lot of computational resources To further illustrate the quadratic congruences, we work three examples with help from Euler's Criterion and from using Excel to do some of the calculations. According to Euler's Criterion, the equationa (mod p) has solutions if and only if only if a2 (mod p). Equivalently, the equation 2a (mod p) has no solutions if and 1 (mod p) So the solvability of the quadratic congruence equation can be translated as a modular exponentiation calculation

5. STEP1: Make 'clen- uatiom aw it is and uane method 16 (mod ) (xt6)2 -36 -I (mod?) (X+ 6)2 52 (mod 7 (2+6)2-3(mo d 7 ) 72= 3(mod 구) Soluutim for 011/--3 So (mod) does not haue. 2 + 3x -4o amu solutiom

13-3a3-13.2 χ t L 3-0 (modig uare method (x-13 )2 -169I3o (modlg (X-13)2-156 (modig (X-13) 4 mod 19 Thun ス2 = 4 mod Ig I9-1 741 (mod 19

STEP4: keep on adding."p, to ‘a, urdu uuger 4+19 23 4t209)s 42 4 t 4 (19) 80 4+3C19) 4tIS(旧)229 ス2-4 mod ig (adding i9 dosnt et ou Solutuon ス2ニ 14a (mod. 19)

STEPI multu 8312-8°nt82 o(mod 23) (x-g) 56 (mod 23) (x-&)*10 (mod23 STEP 3: ス2 10 (mod 23 23- 10 し (mod 23 10 L. (mod 23 nt haue a solutcon ot