. Carnegie Institution of Washington publication. . 242 HISTORY OF THE THEORY OF NUMBERS. [CHAP. VI primes 4k + 1. To find a, b without trial, express p, p' as HI by continued fractions and apply (1) and (P2 + Q2)< = £2 + fJ2} H = tp^Q - G. L. Dirichlet88 used the theory of binary quadratic forms to prove that, if m is a product of powers of n primes 4h + 1, the number of sets of relatively prime solutions x, y of x2 + y* = m is 2'x+2. The number (§ 91) of all sets of solutions is the quadruple of the excess of the number of its divisors 4h + 1 over the number of its divisors 4h + 3. A. Ver
. Carnegie Institution of Washington publication. . 242 HISTORY OF THE THEORY OF NUMBERS. [CHAP. VI primes 4k + 1. To find a, b without trial, express p, p' as HI by continued fractions and apply (1) and (P2 + Q2)< = £2 + fJ2} H = tp^Q - G. L. Dirichlet88 used the theory of binary quadratic forms to prove that, if m is a product of powers of n primes 4h + 1, the number of sets of relatively prime solutions x, y of x2 + y* = m is 2'x+2. The number (§ 91) of all sets of solutions is the quadruple of the excess of the number of its divisors 4h + 1 over the number of its divisors 4h + 3. A. Vermehren,89 to express z3 as a sum of two squares, put z = u -f v; then z3 = u?(u + 3v) + v2(3it + v). He took u + 3y = 4n2, 3u + v = 4m2. F. Unferdinger90 noted that the product of the expansions of (a db bi)m gives (a2 + 62)m = A2 + J52, where A, 5 are known polynomials. He84 had shown that a product P of n sums of two squares can be expressed as a E) in 271"1 ways distinct in general. The same result therefore holds for Pm. G. C. Gerono90a proved that every divisor of a sum of two relatively prime squares is a sum of two relatively prime squares. V. Eugenio91 proved the Lemma24 as follows. Let M divide P2 + Q2, where P is prime to Q, and call P'/Q' the next to the last convergent of the continued fraction for P/Q. Then PQ' - P'Q = d= 1. By (1), M divides (PPf + QQ'Y + 1. Thus M divides AT2 + 1, where N is an integer < M. Express M/N as a continued fraction with an even number of quotients: where n = 2s. Let Mi/Ni, • • • , Mn/Nn = M/N be the successive con- vergents. Then (4) Mi+l -i an-2 + • • • + ai + a Now N2 + 1 = MN'. Thus by (48), -MW' - JVn_i) = N(N - Mn-i). Thus M divides A/" • - Mn_i < M. Hence Jlf n_i = N. Thus (5) equals f, and a = an-i, etc. Hence = l J : " __ _ " ~ But M_! = N8. Thus M/AT = (Jf! + N2a)/MsNs, M = M] + Nl 88 Zahlentheorie, §68, 1863; ed. 2, 1871; ed. 3, 1879; ed. 4, 1894. 89 Die Pythagoraischen Zahlen, Progr
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