By Carl Ludwig Siegel
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Ak be integers, not all zero. For any integer b, there exist integers x1 , . . 4) if and only if b is a multiple of (a1 , . . , ak ). 4) has a solution for every integer b if and only if the numbers a1 , . . , ak are relatively prime. Proof. Let d = (a1 , . . , ak ). 4) is solvable in integers xi , then d divides b since d divides each integer ai . Conversely, if d divides b, then b = dq for some integer q. 4, there exist integers y1 , . . , yk such that a1 y1 + · · · + ak yk = d. Let xi = yi q for i = 1, .
AN = b0 , b1 , . . , bM with ai , bj ∈ Z and aN , bM ≥ 2, then a0 = [x] = b0 . 9. Prove that a0 , a1 , . . , aN , aN +1 = a0 , a1 , . . , aN + 1 . aN +1 10. Let a0 , a1 , . . , aN be a ﬁnite simple continued fraction. Deﬁne p 0 = a0 , p1 = a1 a0 + 1, and pn = an pn−1 + pn−2 for n = 2, . . , N . 3 The Euclidean Algorithm and Continued Fractions 23 and qn = an qn−1 + qn−2 for n = 2, . . , N . Prove that a0 , a1 , . . , an = pn qn for n = 0, 1, . . , N . The continued fraction a0 , a1 , .
AN = a0 + . ··· a1 + a2 + aN The numbers a0 , a1 , . . , aN are called the partial quotients of the continued fraction. For example, 2, 1, 1, 2 = 2 + 1 1+ 1 1+ 12 = 13 . 5 We can write a ﬁnite simple continued fraction as a rational function in the variables a0 , a1 , . . , aN . For example, a0 = a0 , 20 1. Divisibility and Primes a0 , a1 = and a0 , a1 , a2 = a0 a1 + 1 , a1 a0 a1 a2 + a0 + a2 . a1 a2 + 1 If N ≥ 1, then (Exercise 5) a0 , a1 , . . , aN = a0 + 1 . a1 , . . , aN We can use the Euclidean algorithm to write a rational number as a ﬁnite simple continued fraction with integral partial quotients.