The Chinese Remainder Theorem
Chinese Remainder Theorem: If m1, m2, .., mk are pairwise relatively
prime positive integers, and if a1, a2, .., ak are any integers, then the
simultaneous congruences
x ≡ a1 (mod m1), x ≡ a2 (mod m2), ..., x ≡ ak (mod mk)
have a solution, and the solution is unique modulo m, where
m = m1m2⋅⋅⋅ mk.
Proof that a solution exists: To keep the notation simpler, we will
assume k = 4. Note the proof is constructive, i.e., it shows us how to
actually construct a solution.
Our simultaneous congruences are
x ≡ a1
(mod
m
1
)
, x ≡ a2
(mod
m
2
)
, x ≡ a3
(mod
m
3
), x ≡ a4
(mod
m
4
)
.
Our goal is to find integers w1, w2, w3, w4 such that:
value
mod m1
value
mod m2
value
mod m3
value
mod m4
w1 1 0 0 0
w2 0 1 0 0
w3 0 0 1 0
w4 0 0 0 1
Once we have found w1, w2, w3, w4, it is easy to construct x:
x = a1w1 + a2w2 + a3w3 + a4w4.
Moreover, as long as the moduli (m1, m2, m3, m4) remain the same, we
can use the same w1, w2, w3, w4 with any a1, a2, a3, a4.
First define: z1 = m / m1 = m2m3m4
z2 = m / m2 = m1m3m4
z3 = m / m3 = m1m2m4
z4 = m / m4 = m1m2m3
Note that
i) z1 ≡ 0 (mod mj) for j = 2, 3, 4.
ii) gcd(z1,m1) = 1. (If a prime p dividing m1 also divides
z1= m2m3m4, then p divides m2, m3, or m4.)
and likewise for z2, z3, z4.
Next define: y1 ≡ z1–1 (mod m1)
y2 ≡ z2–1 (mod m2)
y3 ≡ z3–1 (mod m3)
y4 ≡ z4–1 (mod m4)
The inverses exist by (ii) above, and we can find them by Euclid’s
extended algorithm. Note that
iii) y1z1 ≡ 0 (mod mj) for j = 2, 3, 4. (Recall z1 ≡ 0 (mod mj) )
iv) y1z1 ≡ 1 (mod m1)
and likewise for y2z2, y3z3, y4z4.
Lastly define: w1 ≡ y1z1 (mod m)
w2 ≡ y2z2 (mod m)
w
3 ≡ y3z3 (mod m)
w
4 ≡ y4z4 (mod m)
Then w1, w2, w3, and w4 have the properties in the table on the
previous page.
