Chinese Remainder Theorem

A First Course in Number Theory

2.4 Chinese Remainder Theorem

6. Prove thatmis prime if and only ifϕ(m) =m−1.

7. Prove thatϕ(m) =ϕ(2m) if and only ifmis odd.

8. Prove that if mdividesn, thenϕ(m) dividesϕ(n).

9. Find all positive integersnsuch thatϕ(n) is not divisible by 4.

10. Find all positive integersnsuch thatϕ(5n) = 5ϕ(n).

11. Letf(n) =ϕ(n)/n. Prove thatϕ(p^k) =ϕ(p) for all primespand all positive integersk.

12. This problem gives an alternative proof of Theorem 2.8. Letm≥1, and letS be the set of fractionsk/mwithk= 0,1, . . . , m−1. Write each fraction in lowest terms:k/m=a/d, wheredis a divisor ofm and (a, d) = 1. For example, 0/m= 0/1. Show that for each divisord ofmthere are exactlyϕ(d) fractionsk/m∈Sthat have denominator dwhen reduced to lowest terms. Deduce that

d|mϕ(d) =m.

13. Let Nm(x) denote the number of positive integers not exceeding x that are relatively prime tom. Prove that

x→∞lim Nm(x)

x = ϕ(m) m .

This result can be expressed as follows: The probability that a random integer is prime tomisϕ(m)/m.

62 2. Congruences

Proof. Ifxis a solution of congruence (2.5), thenx=a+mufor some integer u. Ifxis also a solution of congruence (2.6), then

x=a+mu≡b (modn), that is,

a+mu=b+nv for some integerv. It follows that

a−b=nv−mu≡0 (mod (m, n)).

Conversely, ifa−b≡0 (mod (m, n)), then by Theorem 1.15 there exist integersuandv such that

a−b=nv−mu.

Then

x=a+mu=b+nv is a solution of the two congruences.

An integery is another solution of the congruences if and only if y≡a≡x (modm)

and

y≡b≡x (modn),

that is, if and only ifx−yis a common multiple ofmandn, or, equivalently, x−y is divisible by the least common multiple [m, n]. This completes the proof.2

For example, the system of congruences x ≡ 5 (mod 21), x ≡ 19 (mod 56), has a solution, since

(56,21) = 7 and

19≡5 (mod 7).

The integerxis a solution if there exists an integerusuch that x= 5 + 21u≡19 (mod 56),

that is,

21u≡14 (mod 56),

3u≡2 (mod 8), or

u≡6 (mod 8).

Then

x= 5 + 21u= 5 + 21(6 + 8v) = 131 + 168v

is a solution of the system of congruences for any integerv, and so the set of all solutions is the congruence class 131 + 168Z.

Theorem 2.10 (Chinese remainder theorem) Letk≥2. Ifa1, . . . , ak

are integers andm1, . . . , mk are pairwise relatively prime positive integers, then there exists an integerxsuch that

x≡ai (mod mi) for alli= 1, . . . , k.

If xis any solution of this set of congruences, then the integer y is also a solution if and only if

x≡y (modm₁· · ·m_k).

Proof. We prove the theorem by induction onk. Ifk= 2, then [m1, m2] = m1m2, and this is a special case of Theorem 2.9.

Letk≥3, and assume that the statement is true fork−1 congruences.

Then there exists an integerzsuch thatz≡a_i (mod m_i) fori= 1, . . . , k− 1. Sincem₁, . . . , m_k are pairwise relatively prime integers, we have

(m1· · ·m_k−1, mk) = 1,

and so, by the casek= 2, there exists an integerxsuch that x ≡ z (modm1· · ·mk−1),

x ≡ a_k (mod m_k).

Then

x≡z≡ai (modmi) fori= 1, . . . , k−1.

If y is another solution of the system of k congruences, then x−y is divisible bymi for alli= 1, . . . , k. Sincem1, . . . , mk are pairwise relatively prime, it follows that x−y is divisible by m1· · ·mk. This completes the proof.2

For example, the system of congruences x ≡ 2 (mod 3), x ≡ 3 (mod 5), x ≡ 5 (mod 7), x ≡ 7 (mod 11)

64 2. Congruences

has a solution, since the moduli are pairwise relatively prime. The solution to the first two congruences is the congruence class

x≡8 (mod 15).

The solution to the first three congruences is the congruence class x≡68 (mod 105).

The solution to the four congruences is the congruence class x≡1118 (mod 1155).

There is an important application of the Chinese remainder theorem to the problem of solving diophantine equations of the form

f(x1, . . . , xk)≡0 (modm),

wheref(x1, . . . , xk) is a polynomial with integer coefficients in one or sev- eral variables. This equation is solvable modulo m if there exist integers a1, . . . , ak such that

f(a₁, . . . , a_k)≡0 (modm).

The Chinese remainder theorem allows us to reduce the question of the solvability of this congruence modulomto the special case of prime power modulip^r. For simplicity, we consider polynomials in only one variable.

Theorem 2.11 Let

m=p^r₁¹· · ·p^r_k^k

be the standard factorization of the positive integerm. Letf(x)be a poly- nomial with integral coefficients. The congruence

f(x)≡0 (modm) is solvable if and only if the congruences

f(x)≡0 (mod p^r_iⁱ) are solvable for all i= 1, . . . , k.

Proof. Iff(x)≡0 (modm) has a solution in integers, then there exists an integer asuch thatm dividesf(a). Sincep^r_iⁱ dividesm, it follows that p^r_iⁱ dividesf(a), and so the congruencesf(x)≡0 (mod p^r_iⁱ) are solvable fori= 1, . . . , k.

Conversely, suppose that the congruencesf(x)≡0 (modp^r_iⁱ) are solv- able for i= 1, . . . , k. Then for eachithere exists an integera_i such that

f(ai)≡0 (mod p^r_iⁱ).

Since the prime powersp^r₁¹, . . . , p^r_k^k are pairwise relatively prime, the Chi- nese remainder theorem tells us that there exists an integerasuch that

a≡ai (mod p^r_iⁱ) for alli. Then

f(a)≡f(ai)≡0 (mod p^r_iⁱ)

for all i. Sincef(a) is divisible by each of the prime powersp^r_iⁱ, it is also divisible by their product m, and sof(a)≡0 (modm). This completes the proof.2

For example, consider the congruence

f(x) =x²−34≡0 (mod 495).

Since 495 = 3²·5·11, it suffices to solve the congruences f(x) =x²−34≡x²+ 2≡0 (mod 9), f(x) =x²−34≡x²+ 1≡0 (mod 5), and

f(x) =x²−34≡x²−1≡0 (mod 11).

These congruences have solutions

f(5)≡0 (mod 9), f(2)≡0 (mod 5), and

f(1)≡0 (mod 11).

By the Chinese remainder theorem, there exists an integerasuch that a ≡ 5 (mod 9),

a ≡ 2 (mod 5), a ≡ 1 (mod 11).

Solving these congruences, we obtain

a≡122 (mod 495).

We can check that

f(122) = 122²−34 = 14,850 = 30·495, and so

f(122)≡0 (mod 495).

66 2. Congruences

Exercises

1. Find all solutions of the system of congruences x ≡ 4 (mod 5), x ≡ 5 (mod 6).

2. Find all solutions of the system of congruences x ≡ 5 (mod 12), x ≡ 8 (mod 9).

3. Find all solutions of the system of congruences x ≡5 (mod 12), x ≡8 (mod 10).

4. Find all solutions of the system of congruences 2x ≡1 (mod 5), 3x ≡4 (mod 7).

5. Find all integers that have a remainder of 1 when divided by 3, 5, and 7.

6. Find all integers that have a remainder of 2 when divided by 4 and that have a remainder of 3 when divided by 5.

7. Find all solutions of the congruence

f(x) = 5x³−93≡0 (mod 231).

8. (Bhaskara, sixth century) A basket contains n eggs. If the eggs are removed 2,3,4,5,or 6 at a time, then the number of eggs that remain in the basket is 1,2,3,4, or 5, respectively. If the eggs are removed 7 at a time, then no eggs remain. What is the smallest numbernof eggs that could have been in the basket at the start of this procedure?

Hint:The first condition implies thatn≡1 (mod 2).

9. Letf be a polynomial with integer coefficients. Form≥1, letNf(m) denote the number of pairwise incongruent solutions of f(x) ≡ 0 (modm). Prove that the function Nf(m) is multiplicative, that is, Nf(m1m2) =Nf(m1)Nf(m2) if (m1, m2) = 1.

10. Letm1, . . . , mkbe pairwise relatively prime positive integers andm= m1· · ·mk. Define the map

f : (Z/mZ)^×→(Z/m1Z)^×× · · · ×(Z/mkZ)^× by

f(a+mZ) = (a+m₁Z, . . . , a+m_kZ).

Use the Chinese remainder theorem to show directly that this map is one-to-one and onto.