Skip to main content

Full text of "History Of The Theory Of Numbers - I"

See other formats

Fermat1 stated that (2?+l)/3 has no factors other than 2kp+l if p is an odd prime.
L. Euler2 noted that a4+464 has the factors a2==2a&+2&2.
Euler3 discussed the numbers a for which a2-f 1 is divisible by a prime 4n+l = r2+s2. Let p/q be the convergent preceding r/s in the continued fraction for r/s ; then ps  qr = == 1 . Thus every a is of the form (4n+ l)m== k, where k = pr+qs.
Euler4 gave the 161 integers a< 1500 for which a2+l is a prime, and the cases a = 1, 2, 4, 6, 16, 20, 24, 34 for which a4+l is a prime.
Euler8 proved that, if m is a prime and a, b are relatively prime, a factor of am  bmj not a divisor of a  6, is of the form kn + 1 . If p = kn + 1 is a prime and a =/*=*= pa, then a*  1 is divisible by p. If of1  60n is divisible by a prime p = wn+1, while/ and 0 are not both divisible by p, then am 6m is divisible by p] the converse is true if m and n are relatively prime.
Euler6 proved the related theorems: For q an odd prime, any prime divisor of afl 1, not a divisor of a 1, is of the form 2nq+l. If am  1 is divisible by the prime p=wn+l, we can find integers x, y not divisible by p such that A = axn yn is divisible by p (since the quotient of amxmnymn by A is not divisible by p if x, y are suitably chosen).
Euler7 treated the problem to find all integers a for which a2+l is divisible by a given prime 4n + 1 = p2 +q2. If a2 + b2 is divisible by p2 +q2, there exist integers r, s such that a  pr+qSj b = psqr. We wish 6= =*=!. Hence we take the convergent r/s preceding p/q in the continued fraction for p/q. Thus psqr = ==1, and our answer is a= == (pr+qs). He listed all primes P = 4n+l<2000 expressed as p2-f~#2, and listed all the a's for which a2+l is divisible by P. The table may be used to find all the divisors <a of a given number a2+l. He gave his4 table and tabulated the values a< 1500 for which (a2+l)/k is a prime, for k=2, 5, 10. He tabulated all the divisors of a2+l for a^ 1500.
N. Beguelin8 stated that 2n+l has a trinary divisor l+2p-f-2* only when n = 10, 24, 32, although his examples (p. 249) contradict this statement.
Euler9 gave a factor of 2n=*= 1 for various composite n's.
MDeuvres, 2, 205, letter to Frenicle, Aug. (?), 1640.   Bull. Bibl. St. Sc. Mat. e Fis., 12, 1879, 716.
'Corresp. Math. Phya. (ed., Fuss), I, 1843, p. 145; letter to Goldbach, 1742.
/Wd., 242-3; letter to Goldbach, July 9, 1743.
*Ibid., 588-9, Oct. 28, 1752.    Published, Euler.7
6Novi Comm. Petrop., 1, 1747-8, 20; Comm. Arith. Coll., 1, 57-61, and posthumous paper,
tbtU, 2, 530-5; Opera postuma, I, 1S62, 33-35.   Cf. Euler162 of Ch. VII and the topic
Quadratic Residues in Vol. III.
8Novi Comm. Petrop., 7, 1758-9 (1755), 49; Comm. Arith., 1, 269. 7Novi Comm. Petrop., 9, 1762-3, 99; Comm. Arith., 1, 358-369.    French transl., Sphinx-
Oedipe, 8, 1913, 1-12, 21-26, 64.
Mem. Ac. Berlin, ann<5e 1777, 1779, 255.   Cf. Ch. XV and Henry." 9Posthumous paper, Comm. Arith., 2, 551; Opera postuma. I, 1862, 51.