Question

Problem 2. Find a primitive root for 53. Using this, you can devise a bijection α from the integers modulo 52 to the nonzero integers modulo 53 with the property that α(a + b) = α(a)· α(b) modulo 53. Explain. Does the law of exponents get involved at all? Note: For this to work right, you can think of integers mod 52 as {0, 1, 2, . . . , 51} or as any complete system of residues modulo 52, and similarly nonzero integers mod 53 could be thought of as {1, 2, 3, . . . , 52} or any other reduced system of residues mod 53, that is, not containing anything congruent to 0 mod 53. You could also think of equivalence classes for the congruence relation

Problem 2. Find a primitive root for 53. Using this, you can devise a bijection a from the integers modulo 52 to the nonzero integers modulo 53 with the property that a(a+b) = a(a)·a(b) modulo 53. Explain. Does the "law of exponents” get involved at all? Note: For this to work right, you can think of “integers mod 52” as {0, 1, 2, ...,51} or as any complete system of residues modulo 52, and similarly “nonzero integers mod 53” could be thought of as {1,2,3,...,52} or any other “reduced” system of residues mod 53, that is, not containing anything congruent to 0 mod 53. You could also think of equivalence classes for the congruence relation.

Answer 1

% More than 53 is krime no, hence, 9(53) - 53-1 = 52. therefore, multiplicative group of 7253 = (72531{0})* = 252 wst the map : 232,+) (72531103) * a carb)= a(a) a (6) avod 53. Say { is krimitive root of 53. . 953 = 1 mod 53 and 2 & 1 mod 53 for d<53 in the group (22531 {0},*). ire. 53{=0 mod 52 and d{to mod 52 for d<53 in the group (7252,t), Bence, 52 | 532 = 52|as (52,53) = 1 : -) { =52. Anwen 52 is the primitive Groot for 53.