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Modular Arithmetic Warmup

Modular Arithmetic Warmup. Computing powers What is 3 2 (mod 7)? 3 2 = 9 = 2 (mod 7) What is 3 25 (mod 7)? 3 25 = (3 12 ) 2 ×3 3 12 = (3 6 ) 2 3 6 = (3

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Discrete Logarithms So 25 is a base-7 discrete logarithm of 3 since 3 25 = 3 (mod 7) What is log ? Easy using Wolfram alpha: –log b a = log a/log b But what is a discrete base 123 log of 1 (mod 7)? 57 is an answer since = ×7+1 But how would you ever know? And if the base and the modulus get bigger ….

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Page 1: Modular Arithmetic Warmup. Computing powers What is 3 2 (mod 7)? 3 2 = 9 = 2 (mod 7) What is 3 25 (mod 7)? 3 25 = (3 12 ) 2 ×3 3 12 = (3 6 ) 2 3 6 = (3

Modular Arithmetic Warmup

Page 2: Modular Arithmetic Warmup. Computing powers What is 3 2 (mod 7)? 3 2 = 9 = 2 (mod 7) What is 3 25 (mod 7)? 3 25 = (3 12 ) 2 ×3 3 12 = (3 6 ) 2 3 6 = (3

Computing powers• What is 32 (mod 7)?• 32 = 9 = 2 (mod 7)• What is 325 (mod 7)?• 325 = (312)2×3• 312 = (36)2

• 36 = (33)2

• 33 = 32×3 = 2×3 = 6 (mod 7)• 36 = 62 = 1 (mod 7)• 312 = 12 = 1 (mod 7)• 325 = 12 × 3 = 3 (mod 7)

Page 3: Modular Arithmetic Warmup. Computing powers What is 3 2 (mod 7)? 3 2 = 9 = 2 (mod 7) What is 3 25 (mod 7)? 3 25 = (3 12 ) 2 ×3 3 12 = (3 6 ) 2 3 6 = (3

Discrete Logarithms• So 25 is a base-7 discrete logarithm of 3 since 325 = 3 (mod 7)• What is

log123133226724974042726916099225072978121787999602681264220242808237393822637462751150704898781659019329899261348951831735003?

• Easy using Wolfram alpha: – logba = log a/log b

• But what is a discrete base 123 log of 1 (mod 7)?• 57 is an answer since 12357 =

19032389282006103845157032153282588826857086097323460034686891056260376780393021529271254522717047128465906993118819286×7+1

• But how would you ever know?• And if the base and the modulus get bigger ….