Modular Arithmetic: Fermat’s Theorem,
The Chinese Remainder Theorem, Euler’s Theorem
𝑨 ≡ 𝑩 (𝒎𝒐𝒅 𝒏)
𝑨 − 𝑩 𝒊𝒔 𝒂 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝒐𝒇 𝒏
means:
𝑨=𝑩+𝒌×𝒏
equivalently:
for some integer 𝑘.
General Mod Arithmetic rules:
If
𝑨𝟏 ≡ 𝑩𝟏 (𝒎𝒐𝒅 𝒏)
and 𝑨𝟐 ≡ 𝑩𝟐 (𝒎𝒐𝒅 𝒏) then
𝑨𝟏 + 𝑨𝟐 ≡ 𝑩𝟏 + 𝑩𝟐 (𝒎𝒐𝒅 𝒏),
𝑨𝟏 𝑨𝟐 ≡ 𝑩𝟏 𝑩𝟐 (𝒎𝒐𝒅 𝒏),
𝑨𝒌𝟏 ≡ 𝑩𝒌𝟏 (𝒎𝒐𝒅 𝒏).
1. Mod 5:
21 ≡
31 ≡
41 ≡
20141 ≡
22 ≡
32 ≡
42 ≡
20142 ≡
23 ≡
33 ≡
43 ≡
20143 ≡
24 ≡
34 ≡
44 ≡
20144 ≡
22012 ≡
32012 ≡
42012 ≡
20142012 ≡
22014 ≡
32014 ≡
42014 ≡
20142014 ≡
Conclusion:
𝒂𝟒 ≡ 𝟏
(mod 𝟓)
for all integers 𝒂 ≠ multiple of 5.
2. Mod 7:
21 ≡
31 ≡
41 ≡
20141 ≡
22 ≡
32 ≡
42 ≡
20142 ≡
23 ≡
33 ≡
43 ≡
20143 ≡
26 ≡
36 ≡
46 ≡
20146 ≡
22014 ≡
32014 ≡
42014 ≡
20142014 ≡
Conclusion:
𝒂𝟔 ≡ 𝟏
(mod 𝟕)
for all integers 𝒂 ≠ multiple of 7.
3. Fermat’s Little Theorem:
If 𝒑 is a prime number and 𝒂 is a number not divisible by 𝒑, then
Equivalently,
𝒂𝒑−𝟏 ≡ 𝟏
(mod 𝒑).
𝒂𝒑 ≡ 𝒂
(mod 𝒑).
Hint for proof: Given as many beads as you want in 𝒂 available colours, how many different
types of necklaces of 𝒑 beads can you form which use more than just 1 colour? 𝒑 is prime.
4. Calculate 𝑎) 32014 (mod 11)
𝑏) 999999 (mod 11).
𝑐) 201313 (mod 13)
𝑑) 100100 (mod 31)
Products:
Mod 35:
If we know a number 𝑨 (mod 5) and 𝑨 (mod 7), then we can find 𝑨 (mod 35) uniquely.
5. Use the results from exercises 1 and 2 to fill in the first two rows, and then use these to
calculate the entries in the last row.
22014 ≡
(mod 5)
32014 ≡
(mod 5)
20142014 ≡
(mod 5)
22014 ≡
(mod 7)
32014 ≡
(mod 7)
20142014 ≡
(mod 7)
22014 ≡
(mod 35) 32014 ≡
(mod 35)
20142014 ≡
(mod 35)
Hint: If 𝐴 ≡ a (mod 5) and 𝐴 ≡ b (mod 7) , then 𝐴 = 5𝑥 + 𝑎 ≡ b (mod 7). Use this
equation to find 𝑥 (mod 7). That’s enough to determine 𝐴 (mod 35).
The Chinese Remainder Theorem:
If 𝒎 and 𝒏 are relatively prime (no common factors >1, or equivalently gcd(𝒎, 𝒏)=1 ), then
𝑨 (mod 𝒎) and 𝑨 (mod 𝒏), are enough to uniquely determine 𝑨 (mod 𝒎𝒏).
6. Let 𝑎 be a number not divisible by 5, 7, 11, or 13. Prove the following:
i) 𝑎24 ≡ 1 (mod 35),
ii) 𝑎720 ≡ 1 (mod 1001).
Hint: Use the Chinese Remainder Theorem together with Fermat’s Little Theorem.
In general:
If 𝒏 = 𝒑𝟏 𝒑𝟐 ⋯ 𝒑𝒔 is a product of distinct primes and if 𝒂 is not divisible by any of these,
then
𝒂(𝒑𝟏 −𝟏)(𝒑𝟐 −𝟏)⋯(𝒑𝒔−𝟏) ≡ 𝟏 (mod 𝒑𝟏 𝒑𝟐 ⋯ 𝒑𝒔 ).
Powers:
7. a) If 𝒂 ≡ 𝒃 (mod 5), prove the following:
i) 𝒂𝟓 ≡ 𝒃𝟓 (mod 25),
ii) 𝒂𝟐𝟓 ≡ 𝒃𝟐𝟓 (mod 125),
iii) 𝒂𝟏𝟐𝟓 ≡ 𝒃𝟏𝟐𝟓 (mod 625).
b) If 𝒂 is not a multiple of 5, prove the following:
i) 𝒂𝟐𝟎 ≡ 𝟏 (mod 25),
ii) 𝒂𝟏𝟎𝟎 ≡ 𝟏 (mod 125),
iii) 𝒂𝟓𝟎𝟎 ≡ 𝟏 (mod 625).
Hint: a) i) Use 𝑎5 − 𝑏 5 = (𝑎 − 𝑏)(𝑎4 + 𝑎3 𝑏 + 𝑎2 𝑏 2 + 𝑎𝑏 3 + 𝑏 4 ) and calculate each of
these factors mod 5. ii) Use 𝑎25 = (𝑎5 )5 . b) Use a) and Fermat’s Little Theorem.
In general:
Let 𝒑 be a prime number. If 𝒂 ≡ 𝒃 (mod 𝒑), then:
i) 𝒂𝒑 ≡ 𝒃𝒑 (mod 𝒑𝟐 ),
𝟐
𝟐
𝒌
ii) 𝒂𝒑 ≡ 𝒃𝒑 (mod 𝒑𝟑 ),
𝒌
iii) 𝒂𝒑 ≡ 𝒃𝒑 (mod 𝒑𝒌+𝟏 ).
b) If 𝒑 is a prime number and 𝒂 is a number not divisible by 𝒑, then
𝒂(𝒑−𝟏)𝒑
𝑘
Hint: a) iii): Induction, using 𝑎𝑝 = (𝑎𝑝
𝑘−1
𝒏−𝟏
≡𝟏
(mod 𝒑𝒏 ).
𝑝
) .
Euler’s Theorem:
𝒌
𝒌
𝒌
If 𝒏 = 𝒑𝟏𝟏 𝒑𝟐𝟐 ⋯ 𝒑𝒔 𝒔 is the prime factorization of 𝒏, we define Euler’s number of 𝒏 by:
𝒌 −𝟏
𝒌 −𝟏 𝒌𝟐 −𝟏
𝒑𝟐
⋯ 𝒑𝒔 𝒔 .
𝝋(𝒏) = (𝒑𝟏 − 𝟏)(𝒑𝟐 − 𝟏) ⋯ (𝒑𝒔 − 𝟏)𝒑𝟏𝟏
If 𝒂 is not divisible by any of these primes, then
𝒌𝟏 −𝟏 𝒌𝟐
𝒌
𝒑𝟐 ⋯𝒑𝒔 𝒔
𝒂(𝒑𝟏 −𝟏)(𝒑𝟐 −𝟏)⋯(𝒑𝒔−𝟏)𝒑𝟏
≡𝟏
(mod 𝒏).
Exercise:
𝝋(𝒏) = how many positive numbers smaller than 𝒏 are relatively prime with 𝒏.