By Peter Barlow

ISBN-10: 1429700467

ISBN-13: 9781429700467

Barlow P. An easy research of the idea of numbers (Cornell collage Library, 1811)(ISBN 1429700467)

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**Example text**

The solution of the equation ax + c = by where a, b, c are given integers with a > b and gcd(a, 6) = 1, is called Kuttaka by Indian mathematicians [4, 27, 89]. It was told that the Indian mathematician Aryabhata (c. D. 476) was able to solve the two-congruence Chinese remainder problem with the help of continued division (Euclidean algorithm) (see Aryabhatiya by Aryabhata, English translation by K. S. Shukla, 1979, Delhi). 59] an Islamic scholar Ibn al-Haitam (about 1000) gave two methods to find a number, divisible by 7, which has the remainder 1 when divided by 2, 3, 4, 5, or 6.

This proves the uniqueness. • Similarly, one can prove the following Chinese Remainder Theorem for polynomials. 2 Let m,-(a;) be pairwise relatively prime polynomials of F\x\. , n — 1 has exactly one solution modulo m(x) = m0(x)mi(x) • • • m n _ 1 (x). The above two versions of the CRT are only of concern for our applications in later chapters. However, it could be helpful to interpret the CRT from a ring-theoretic point of view. ,an + bn). 4. A GENERALIZED CRA 25 Let R be a commutative ring with identity, and / be an ideal of R.

XW)|. " " [dA) To ensure [ F ^ 1 ' , . . ,x^) with a, < x^ < bi, it is necessary that m > M2 + 1 . For example, let F{x) = x, where 0 < x < 6, then M 2 = 6 with respect to this arithmetic expression. If we choose m = M 2 , then F{b) mod m = F(0) mod m = 0, but F(6) = 6 ^ 0 . However, the condition m > M 2 + 1 is not enough to ensure that different 36 CHAPTER 3. ,xW) have different images. For example, let F(x) = x, where —6 < x < b, then M2 = b. Choose m = b+l, then F(b) mod m = F( — 1) mod m = b, but 6 ^ — 1 .