Question

(a) Solve the simultaneous congruences p = 1 (mod x – 3), p = 7 (mod x – 5). (b) Find the total number of monic irreducible polynomials of degree 5 in Fr[c]. (c) Find a primitive root modulo 52020. (Make sure to justify your answer.) (d) Determine the total number of primitive roots modulo 52020.

Answer 1

3 { 1,-1,0} such that n=1 0 1 / E uld) F dus divisor of n we know thet, mobious femction Mi N Min) = if n has a square fecter n has or distinct prime feetex anal Number of monic iruducible polynomial of degreen over zp is (old) yin) - din if 4 (-1) 4 also we know that field of order of le z bu is isomorphic Zz Here n = 5 anel p= i d=1,5 Eld) 4 (5) = 15 E I E uld) = $ [cena + 415)7] (-i)'7] 75 + ull) = 1 1115) = -1 )) & [ 75 - 7 3 [77 - 1] = 1 / 1 7 (2401-1) 16800 5 415) = 3 360 which is the total number of irreducible. palynomial of degro 5 in Zz [x]

which is the total number vwot of modulo 52020. Now 2 2019) 2 $ (4 x 52019) (2 5 2019 2² 5 (1-1) (1-4) 42 = 2² 2019 3 2018 2 5 ở (+ x^^! ) $($(52020)] 3 2018. 2 5 € Since 3 (4 x 5 52019) i (mod 52020) ; ged (3,5 2020)=) I (mod m) uy ged(a,m)=1} . 3 is primitive owot { a dem =

J Grinen pel mod (x-3) and pe 7 mod1X-5) (-3)/(P-1) and x-5 l p-7 (x - 3) / (P-1) 7 a paritine integen ki such theit (P-1) = K, (x-3) p = P = k1. (x-3) + 1 - 2 . (X-5) [(P-7) 7 a positive integer K2 such that (P-7) - Kz (2-5) → P= Xy (x-7)+7 from o o and K, (- 3) + 1 = Kz (X-5)+7 23 xk, -3K, Kzx -s ka + 6 kk, - xk2 3k, - SK2+6 3K, -3kz - 2kg +6 X(ki - K₂) x 3 - 2kz KiKz + 6 K, - К. a bom we know that if a is perimitive root with suspect to n then = 1 (modo); god (9,0) = 1 and of primitive doot with ouspect to n {$($ (n)] Here n= 52020 $ (52020) 2020 5 Now 11-$) 4 $20 2020 2020 $652020) 4 x 5 2019