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The Chinese Remainder Theorem

Theorem.Letmandnbe two relatively

prime positive integers. Letaandbbe any two integers. Then the two congruences x≡a(modm) x≡b(modn) have common solutions. Any two common so- lutions are congruent modulomn.

The proof gives an algorithm for computing

the common solution. 1

Proof:Since gcd(m,n) = 1, the Extended

Euclidean Algorithm provided integerscand

dwith mc+nd= 1.

Thenc≡m-1(modn) andd≡n-1(modm).

Letx0=mcb+nda. Then

0≡nda≡1·a≡a(modm)

and x

0≡mcb≡1·b≡b(modn).

Thus there is a common solutionx0.

Ifx1is another common solution, then

m|(x0-x1) andn|(x0-x1), somn|(x0-x1) because gcd(m,n) = 1. 2

Example:Solve the system of congruences

x≡1 (mod 7) x≡3 (mod 10).

Note that the hypotheses of the Chinese re-

mainder theorem are satisfied in this example because 7 and 10 are relatively prime.

We havem= 7,n= 10,a= 1,b= 3 and

mn= 70. The extended Euclidean algorithm gives 7(3)+10(-2) = 1, soc= 3 andd=-2.

Then the solution is

x≡x0=mcb+nda= 7(3)3 + 10(-2)1 = = 63-20 = 43 (mod 70). 3

Solvingx2≡a(modn)

We have said nothing (so far) about whether

one can solvex2≡a(modn) whennis a composite number.

We have also said nothing abouthowto solve

it if it has a solution.

There are probabilistic polynomial time algo-

rithms (Tonelli and Cipolla) to compute square roots of QR"s modp, wherepis prime. They work well for numbers of hundreds of digits, but are too complicated to present here. 4 Recall Euler"s Criterion.Theorem. (Euler"s Criterion.) For primep >

2 and 0< a < p,

a (p-1)/2≡1 ifais a QR modpand a (p-1)/2≡ -1 ifais a QNR modp.

Here is a simple algorithm that finds square

roots of QR"s modulo any primep≡3 (mod 4), that is, it works for half of the primes.

Ifp≡3 (mod 4), then the solutions to

2≡a(modp) arex1≡a(p+1)/4(modp)

andx2=p-x1.

To see that this works, note that

x 2 sincea(p-1)/2≡+1 (modp) by Euler"s

Criterion and the fact thatais a QR modp.

Now I will tell you how to solvex2≡a(modn)

whenn=pqis the product of two primes p≡q≡3 (mod 4), an important special case.

Separately solvey2≡a(modp), with solu-

tionsy1andy2, andz2≡a(modq), with solutionsz1andz2. Then use the CRT four times to solve the four systems x≡yi(modp)x≡zj(modq) fori= 1,2;j= 1,2.This will producefour different roots tox2≡a(modn). 6

Example.Find all four square roots of 11

modulo 133.

Factor 133 = 7·19. We must first solvex2≡

11 (modp) forp= 7 and forp= 19.

11 mod 7 = 4, which happens to be 2

2. So the

solution tox2≡11 (mod 7) isx≡ ±2 (mod 7), orx≡2 or 5 (mod 7).

11 mod 19 = 11, so we use exponentiation:

quotesdbs_dbs3.pdfusesText_6

[PDF] The Chinese Remainder Theorem Theorem. Let m and n be two

The Chinese Remainder Theorem

Theorem.Letmandnbe two relatively

The proof gives an algorithm for computing

Proof:Since gcd(m,n) = 1, the Extended

Euclidean Algorithm provided integerscand

Thenc≡m-1(modn) andd≡n-1(modm).

Letx0=mcb+nda. Then

0≡nda≡1·a≡a(modm)

0≡mcb≡1·b≡b(modn).

Thus there is a common solutionx0.

Ifx1is another common solution, then

Example:Solve the system of congruences

Note that the hypotheses of the Chinese re-

We havem= 7,n= 10,a= 1,b= 3 and

Then the solution is

Solvingx2≡a(modn)

We have said nothing (so far) about whether

We have also said nothing abouthowto solve

There are probabilistic polynomial time algo-

2 and 0< a < p,

Here is a simple algorithm that finds square

Ifp≡3 (mod 4), then the solutions to

2≡a(modp) arex1≡a(p+1)/4(modp)

To see that this works, note that

Criterion and the fact thatais a QR modp.

Now I will tell you how to solvex2≡a(modn)

Separately solvey2≡a(modp), with solu-

Example.Find all four square roots of 11

Factor 133 = 7·19. We must first solvex2≡

11 (modp) forp= 7 and forp= 19.

11 mod 7 = 4, which happens to be 2

2. So the

11 mod 19 = 11, so we use exponentiation: