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Congruence Modulo
Moksh Kumar
August 2024
1 Introduction
Congruence Modulo is a mathematical tool that is used for the following pur- poses:
- To find the remainder of high degree exponents divided by a number
- To eliminate the quotient term
2 Notation
a ≡ b (mod n)
where ’a’, ’b’ and ’n’ are integers and n ̸= 0 and 0 ≤ b < n
since ’a’ and ’b’ are congruent or equivalent the notation may also be written as: b ≡ a (mod n) It is read as:
- ’a’ divided by ’n’ gives remainder ’b’
- ’a’ mod ’b’ (in general language )
For Example: 12 ≡ 2 (mod 5)
Or
2 ≡ 12 (mod 5) The basic notation of congruence modulo can also be expressed in the eu- clidean format such as:
a ≡ b (mod n) ⇔ a = nq + b
Where ’q’ is any integer which satisfies the relation between ’a’, ’b’ and ’n’ and 0 ≤ b < n
For Example: 12 ≡ 2 (mod 5) ⇔ 12 = 5q + 2
Where q = 2 satisfies the above Note: Here ’b’ the remainder is generally used as a positive integer but for convenience ’b’ is also used as a negative integer where −n < b < n Like 12 ≡ 5 (mod 7) ⇔ 12 ≡ − 7 (mod 7)
3 Properties
- a ≡ a (mod n)
- If a ≡ b (mod n) and b ≡ c (mod n) then a ≡ c (mod n)
- If a ≡ b (mod n) and c ≡ d (mod n) then a + c ≡ b + d (mod n) Proof: a ≡ b (mod n) ⇔ a = nq 1 + b c ≡ d (mod n) ⇔ c = nq 2 + d where q 1 and q 2 both are integers Adding both equations we get
a + c = n(q 1 + q 2 ) + (b + d)
Which can be written as
a + c ≡ b + d (mod n)
Hence Proved
- If a ≡ b (mod n) and c ≡ d (mod n) then a − c ≡ b − d (mod n) Proof: a ≡ b (mod n) ⇔ a = nq 1 + b c ≡ d (mod n) ⇔ c = nq 2 + d where q 1 and q 2 both are integers Subtracting both equations we get
a − c = n(q 1 − q 2 ) + (b − d)
Which can be written as
a − c ≡ b − d (mod n)
Hence Proved
f (b) =
X^ n
k=
λkbk^ = λ 0 b^0 + λ 1 b^1 + ... + λn− 1 bn−^1 + λnbn
We know that a ≡ b (mod n) Therefore λmam^ ≡ λmbm^ (mod n) And if weighted Xn
m=
λkak^ ≡
X^ n
m=
λkbk^ (mod n)
putting m = k Xn
k=
λkak^ ≡
X^ n
k=
λkbk^ (mod n)
Therefore f (a) ≡ f (b) (mod n)
Hence Proved
Example 1: Show that 2^55 + 1 is divisible by 11 Solution: 25 = 32 ≡ (−1) (mod 11) 255 = (2^5 )^11 ≡ (−1)^11 ≡ −1 (mod 11) So, 2^55 + 1 ≡ 0 (mod 11) Therefore, it is a multiple of 11 Example 2: Find sum of all the integers n such that 1 ≤ n ≤ 1998 and that 60 divides n^3 + 30n^2 + 100n Solution: If 60 = 3 · 4 · 5 and 4| 100 n then 4 should divide n^3 + 30n^2 , i.e., 4 should divide n^2 (n + 30). This implies that n is even, i.e., 2|n As 5|(30n^2 + 100n), 5 should divide n^3. Hence 5 should divide n As 3| 30 n^2 , then 3 should divide n^3 + 100n, i.e., 3 should divide n(n^2 + 100n) = n(n^2 + 1 + 99) If n ≡ ±1 (mod 3), n^2 ≡ 1 (mod 3), and n^2 + 1 ≡ 2 (mod 3), so neither of n^2 + 1 + 99 and n are divisible by 3 From all three points, we find that n must be a multiple of 2 · 3 · 5 = 30. So, we should find the sum of all multiples of 30 less than 1998 Sn = 30 + 60 + ... + 1980 = 30(1 + 2 + ... + 66) = 66330
4 Some Important Theorems
- Euler Totient Function/Theorem:
aϕ(n)^ ≡ 1 (mod n)
Where ϕ(n) represents the number of integers less than or equal to n that are co prime to n ϕ(n) = n · Πp|n
p
Note: Totient function is multiplicative, ϕ(mn) = ϕ(m) · ϕ(n)
- Carmichael function/Theorem:
aλ(n)^ ≡ 1 (mod n)
Where λ(n) represents the smallest number which satisfy the above mod- ulo for every integer a, where gcd(n, a) = 1 If n = 2α^ with α ≥ 3 then
λ(n) =
ϕ(n) 2 If n = pα^ with p ≥ 3 then λ(n) = ϕ(n) If n = 2α^ with α < 3 then
λ(n) = ϕ(n)
If n = Πmi=1pα i iwhere αi ≥ 3 then
lcm(λ(pα 1 1 ), ..., λ(pα mm ))
ap^ ≡ a (mod p)
Where p is any prime number Also can be written as
ap−^1 ≡ 1 (mod p)
(n − 1)! ≡ − 1 (mod n)
Where n ≥ 2 and n is a natural number
- Chinese Remainder Theorem:
x ≡ a 1 (mod n 1 )
x ≡ a 2 (mod n 2 ) . . . x ≡ ar (mod nr )
(q − r)(q + r) = 8d
d being odd =⇒ y and z both are even =⇒ either (q − r) or (q + r) is divisible by 4 If q − p is divisible by 4 that means (q + r) must also be divisible by 4 as
q + r = (q − r) + 2r
Similarly if (q + r) is divisible by 4 that means (q − r) is also divisible by 4 as
(q − r) = (q + r) − 2 r
Thus 16 |(q − r)(q + r)
Thus 16 | 8 d
but d being an odd number above statement is not satisfied hence being a contradiction. Hence, there is no such d that can make ab − 1 a perfect square Problem 2: Find all integer values of ’a’ such that the quadratic expressions (x + a)(x + 1991) + 1 can be factored as (x + b)(x + c), where b and c are integers [RMO 1991] Solution: we have (x + a)(x + 1991) and by vieta we get,
1991 + a = b + c
1991 a + 1 = bc
upon observation, (b − c)^2 = (b + c)^2 − 4 bc (b − c)^2 = (1991 + a)^2 − 4(1991a + 1) (b − c)^2 = (1991 + a)^2 − 4
or (1991 + a)^2 − (b − c)^2 = 4
Since 4 is even that means (1991 + a)^2 and (b − c)^2 both must be either odd or even If (1991 + a)^2 and (b − c)^2 are both even then the only form they can be is 4k =⇒ (1991 + a) and (b − c) both must be of form 2m where both k and are integers Therefore (2m 1 )^2 − (2m 2 )^2 = 4 m^21 − m^22 = 1
since m is an integer, only solution is (1, 0) hence (1991 + a)^2 = 4
(1991 + a) = ± 2
values of a are 1989 and 1993 and (b − c)^2 = 0 b = c
Hence, values of a that satisfies are 1993 and 1989 with b = c
