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42 HlSTORT OF THE THEORY OF NUMBERS. [CHAP. I
then A:a = 3:5; he took A=3J5, 32J3, 33B, but found no solution. For p = 5, Q = 41, we have # = 251, 38A = 21a; set A = 495, whence 3-576 = 38-75, where b is the sum of the divisors of B; set £ = 9C, whence C:c = 13:14, C = 13, yielding the amicable numbers 5-41A, 25LA, where A = 32-7213 = 5733 [the pair VII in Euler's362 list and (7) in the table below]. Again, to make -A/a = 3/8, set A-3B, whence a = 46 and the condition is fc = 2B, whence Bis a perfect number prune to 3. Using 5 = 28, we get A = 84. For use in such questions, Kraft gave a table of the sum of the divisors of each number^ 150. He quoted the rule of Descartes.
L. Euler364 obtained, in addition to two special pairs, 62 pairs [including two false pairs] of amicable numbers of the type aw, an, in which the common factor a is relatively prune to both m and n. He wrote j m for the sum of all the divisors of m. The conditions are therefore
£m~Jn, j a-J m~a(m+ri).
If m and n are both primes, then m=n and we have a repeated perfect number. Euler treated five problems.
(1) Euler's problem 1 is to find amicable numbers apq, ar, where p, q, r, are distinct primes not dividing the given number a. From the first condition we have r = xy — 1 , where z = p+l, y = q+l. From the second,
Let a/(2a — J a) equal b/c, a fraction in its lowest terms. Then
Thus x and y are to be found by expressing b2 as a product of two factors, increasing each by b} and dividing the results by c.
(10 First, take a = 2n. Then 6 = 2", c = l,z, y = 2n*A+2n. Letn-& = m. Then
p = 2m(22k+2k)-l, 3 = 2m(l+2*)-l, r = 22m(22h+1+23k+2k)-l. When these three are primes, 2m+kpq and 2m+*r are amicable. Euler noted that the rule communicated by Descartes to van Schooten is obtained by taking k—1, and stated that 1, 3, 6 are the only values ^8 of m which yield amicable numbers (above355). For fc = 2 or 4, Euler remarked that r is divisible by 3; for fc = 3, m<6, and for k = 5, m^2, p} q} or r is composite.
(12) Take a = 2*/, where /=2n+1+e is a prime. Then 2a-Ja = e-fl. If e+1 divides a, we have c = l. Set e-fl=2fc, k^m, n = m-f k. Then /=2fc(2-H+l)-l, a = 2"l+A:/, b = 2mf, b2=(x-b)(y-b).
For fc = l, /=2m+2-hl is to be a prime, whence m-f 2 is a power of 2. If 7?z = 0, b=/=5, and either x = y, p = q; or x, y = 6, 30; p} q = 5, 29, whereas p and q are to be distinct and prime to 10. If m = 2, /= 17, 682 is to be resolved into distinct even factors; in the four resulting cases, p, q, r are
3MDe numcris amicabilibus, Opuscula varii argument!, 2, 1750, 23-107, Berlin; Comm. Arith., 1, 1849, 102-145. French transl. in Sphinx-Oedipe, Nancy, 1, 1906-7, Supplement