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Modular Arithmetic
Modular arithmetic is a way of systematically ignoring differences involving a multiple of an integer. If n is an integer, two integers are equal mod n if they differ by a multiple of n; it is as if multiples of n are “set equal to 0”.
Definition. Let n, x, and y be integers. x is congruent to y mod n if n | x − y. Notation:
x = y (mod n).
Remarks. n | x − y is equivalent to the following statements:
(a) n | y − x.
(b) x = y + jn for some j ∈ Z.
(c) y = x + kn for some k ∈ Z.
I’ll often use any of these four statements as the definition of x = y (mod n).
A lot of people like to write “x ∼= y (mod n)” instead of “x = y (mod n)”. I don’t think there’s any harm in using an ordinary equal sign, since the “ (mod n)” makes the meaning clear. It’s also a bit shorter to write.
Example. (Examples of congruences with numbers) (a) Demonstrate that 7 = 1 (mod 6) and 57 = −13 (mod 7).
(b) Express “x is even” and “x is odd” in terms of congruences.
(c) What does x = 0 (mod n) means in terms of divisibility?
(a) 7 = 1 (mod 6) , since 6 | 7 − 1.
57 = −13 (mod 7) , since 7 | 57 − (−13).
(b) x is even if and only if x = 0 (mod 2) and x is odd if and only if x = 1 (mod 2). (c) x = 0 (mod n) if and only if n | x. Thus, congruences provide a convenient notation for dealing with divisibility relations.
The following proposition says that you can work with modular equations in many of the ways that you work with ordinary equations.
Proposition. Let n ∈ Z.
(a) If a = b (mod n) and c = d (mod n), then
a + c = b + d (mod n).
(b) If a = b (mod n) and c = d (mod n), then
ac = bd (mod n).
(c) If a = b (mod n), then ac = bc (mod n).
Proof. Two ideas for these kinds of proofs:
- You can often prove statements about congruences by reducing them to statements about divisibility.
- You can often prove statements about divisibility by reducing them to (ordinary) equations.
(a) Suppose a = b (mod n) and c = d (mod n). a = b (mod n) means n | a − b and c = d (mod n) means n | c − d. By properties of divisibility,
n | (a − b) + (c − d) = (a + c) − (b + d).
Therefore, a + c = b + d (mod n).
(b) Suppose a = b (mod n) and c = d (mod n). a = b (mod n) means n | a − b, which means a − b = jn for some j ∈ Z. c = d (mod n) means n | c − d, which means c − d = kn for some k ∈ Z. Thus, a = b + jn, c = d + kn, and hence
ac = (b + jn)(d + kn) = bd + bkn + djn + jkn^2 = bd + n(bk + dj + jkn).
This gives ac − bd = n(bk + dj + jkn), so n | ac − bd, and hence ac = bd (mod n).
(c) Suppose a = b (mod n). This means that n | a − b. By properties of divisibility,
n | (a − b)c = ac − bc.
Therefore, ac = bc (mod n).
Example. (Solving a congruence) Solve 3x + 4 = 2x + 8 (mod 9).
In this case, I’ll solve the modular equation by adding or subtracting the same thing from both sides.
3 x + 4 = 2 x + 8 (mod 9) − 4 = 4 (mod 9) 3 x = 2 x + 4 (mod 9) − 2 x = 2 x (mod 9) x = 4 (mod 9)
The solution is x = 4 (mod 9).
Example. Reduce 497 · 498 · 499 (mod 500) to a number in the range { 0 , 1 ,... 499 }, doing the computation by hand.
Note that
497 = −3 (mod 500) , 498 = −2 (mod 500) , 499 = −1 (mod 500).
So 497 · 498 · 499 = (−3)(−2)(−1) = −6 = 494 (mod 500).
The next result says that congruence mod n is an equivalence relation.
Z 3 is called the cyclic group of order 3. The “cyclic” nature of Z 3 can be visualized by arranging the integers in a spiral, with each congruence class on a ray.
0
1
2 3
4
5
6
7
8
When you do arithmetic in Z 3 , it is as if you count in a circle: 0, 1, 2, then back to 0 again. You can form other cyclic groups in an analogous way. For example,
Z 6 = { 0 , 1 , 2 , 3 , 4 , 5 }.
You can do arithmetic in Zn by adding and multiplying as usual, but reducing the results mod n.
Example. (Operation tables for Z 3 ) Construct addition and multiplication tables for Z 3.
For example, as integers 2 + 2 = 4. I divide 4 by the modulus 3 and get a remainder of 1. Hence, 2 + 2 = 1. Likewise, 2 · 2 = 4 = 1 in Z 3.
Example. (Equations in Zn) Find 6 · 7 in Z 11 , 13 + 19 in Z21, and −8 in Z 17.
6 · 7 = 9 in Z 11. 13 + 19 = 11 in Z 21. −8 = 9 in Z 17. −8 means the additive inverse of 8. The last statement is just another way of saying −8 = 9 (mod 17).
Example. (Using modular arithmetic in a divisibility proof) Prove that if n is an integer, then 2 n^2 + 3n + 2 is not divisible by 5.
Every integer n is congruent to one of 0, 1, 2, 3, or 4 mod 5. Therefore, I have 5 cases. In each case, I want to show that 2n^2 + 3n + 2 is not divisible by 5 — or to say it in terms of congruences, I want to show that 2n^2 + 3n + 2 6 = 0 (mod 5). I set n = 0, 1 , 2 , 3 , 4 (mod 5) and “substitute” the value into 2n^2 + 3n + 2. This substitution is justified by the properties of congruences I discussed above. For example, if n = 3 (mod 5), then
n · n = 3 · 3 (mod 5) n^2 = 9 = 4 (mod 5) 2 · n^2 = 2 · 4 (mod 5) 2 n^2 = 8 = 3 (mod 5)
Likewise, 3n = 3 · 3 = 9 = 4 (mod 5). So
2 n^2 + 3n + 2 = 3 + 4 + 2 = 9 = 4 (mod 5).
Essentially, I can plug n = 3 into 2n^2 + 3n + 2, then reduce the result mod 5 to one of 0, 1, 2, 3, or 4. Continuing in this way, I get the following table:
n (mod 5) 0 1 2 3 4 2 n^2 + 3n + 2 (mod 5) 2 2 1 4 1
In all five cases, 2n^2 + 3n + 2 6 = 0 (mod 5). Therefore, 2n^2 + 3n + 2 is never divisible by 5.
I showed earlier how to use algebraic operations to solve simple modular equations. How would you solve something like this: 6 x = 13 (mod 25)? I’d like to divide both sides by 6, but I only know how to add and multiply. I can subtract, but that’s because I can add additive inverses. Well, division is multiplication by the multiplicative inverse; what is a multiplicative inverse mod 25?
Definition. Let a, b ∈ Zn. a and b are multiplicative inverses if ab = 1 (mod n) (or ab = 1 in Zn). If a is the multiplicative inverse of b, you can write a = b−^1. (You don’t write “
b
” unless you’re in a number system like the rational numbers where fractions are in
use.)
Example. (Modular multiplicative inverses) (a) Prove that 6 and 2 are multiplicative inverses mod 11.
(b) Show that 8 does not have a multiplicative inverse mod 12.
(a) 6 · 2 = 1 (mod 11).
(b) One tedious way is to take cases:
n 0 1 2 3 4 5 8 n (mod 12) 0 8 4 0 8 4 n 6 7 8 9 10 11 8 n (mod 12) 0 8 4 0 8 4
Thus, 1 = 3 · 52 + (−5) · 31.
In Z 52 , 52 = 0 and −5 = 47. The equation says 1 = 47 · 31. Thus, 47 is the multiplicative inverse of 31 in Z 52.
Theorem. If (a, n) = 1, then the following equation has a unique solution:
ax = b in Zn.
Proof. If (a, n) = 1, then a has a multiplicative inverse a−^1 in Zn. Thus, aa−^1 = 1 in Zn. First, this means that x = a−^1 b is a solution, since
a(a−^1 b) = (aa−^1 )b = 1 · b = b.
Second, if x′^ is another solution, then ax′^ = b. Multiplying both sides by a−^1 , I get
a−^1 ax′^ = a−^1 b, x′^ = a−^1 b.
That is, x′^ = x. This means the solution is unique.
Example. (Solving modular equations using modular inverses) Solve
13 x = 12 (mod 15).
There is a solution, since (13, 15) = 1. I need to find a multiplicative inverse for 13 mod 15.
The Extended Euclidean Algorithm says that
(−6)(15) + (7)(13) = 1.
Hence, 7 · 13 = 1 (mod 15), i.e. 7 is the multiplicative inverse of 13 mod 15. Multiply the original equation by 7:
7 · 13 x = 7 · 12 (mod 15) , x = 84 = 9 (mod 15).
Proposition. Suppose ac = bc (mod n).
Then a = b
mod n (n, c)
Proof. I have
ac = bc (mod n)
a c (n, c)
= b c (n, c)
mod n (n, c)
a
c (n, c)
− b
c (n, c)
= k ·
n (n, c)
for some k ∈ Z
c (n, c)
(a − b) = k ·
n (n, c)
(Note that (n, c) | c and (n, c) | n, so c (n, c)
and n (n, c)
are actually integers.) Now n (n, c)
divides c (n, c)
(a − b), but ( n (n, c)
c (n, c)
By Euclid’s lemma,
n (n, c)
| a − b. Hence,
a = b
mod
n (n, c)
I can use the preceding result to solve some congruences when I can’t immediately use modular inversion.
Example. Solve 12 x = 30 (mod 34).
Since (12, 34) = 2 6 = 1, 12 doesn’t have a multiplicative inverse mod 34. I’ll use the preceding result. I “cancel” a factor of 6 from 12x and 30, and divide the modulus 34 by (6, 34) = 2:
12 x = 30 (mod 34) 6 · 2 x = 6 · 5 (mod 34) 2 x = 5 (mod 17) 9 · 2 x = 9 · 5 (mod 17) x = 45 = 11 (mod 17)
Since the original congruence was mod 34, I must find all numbers in { 0 , 1 , 2 ,... 33 } which satisfy x = 11 (mod 17). One is obviously 11. Adding 17, I find that 11 + 17 = 28 also works. (Adding 17 again takes me out of the set { 0 , 1 , 2 ,... 33 }.) The solutions are x = 11 (mod 17) and x = 28 (mod 17).
©^ c2018 by Bruce Ikenaga 8
