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

CHAP. I] PERFECT, MULTIPLY PERFECT, AND AMICABLE NUMBERS.         43
not all prime.   In the next case m=6, /=257, Euler examined only the case* x 6 = 25-257, finding q composite.
For k = 2, Euler excluded w== 1, 3 [m =4 is easily excluded].
(13) For k^n in (12), c = 2m, where ra = /c  n.   Then
must be a prime.   Thus we must take as the factors of 62
2mz-& = l,          2my-b = b2,
whence x = 2n+ 2n+1~m, y = bx.   If m = 1, one of
/=2n+2-l,          p=2n+1-l
has the factor 3 and yet must be a prime; hence n = l, g = 27. If m = 2, Euler treated the cases n^5 and found (for n = 2) the pair (4) of the table. [For 6^n^l7, / or p is composite.] For m odd and >1, / or p has the factor 3. For m = 4, n^ 17, no solution results.
(14) For a = 2n(g l)(h  1), where the last two factors are prime, set d = 2a-a.   Then
Euler treated the cases n^3, <2=4, 8, 16, finding only the pair (9).
(15) Special odd values of a led (56-65) to seven pairs (5)-(8), (11)-(13). The cases a = 33-5, 32-72-13-19 were unfruitful.
(2) Euler's problem 2 is to find amicable numbers apq, ars, where p, q, r, s are distinct primes not dividing the given number a. Since fp'fq = Jr-Js, we may set
p-ax  1,          g=/3?/~l,          r=/3o;-l,          s = ay  1.
We set Ja:a = 2bc:b, where Z> and c are relatively prime.   The second condition fa-fpq = a(pq-\-rs) gives
cap Multiply it by ca/3.   Then
Given a, /3, a and hence 6, c, we are to express the second member as a product of two factors and then find x, y.
For a=l, 0 = 3, a = 2n, Euler obtained the pairs (a), (28). For a = 2, 0 = 3, a = 32-5-13, he got (32); fora=l, 0 = 4, a = 33-5, (30). The ratio a:/3 may be more complex, as 5 :21 or 1 :102, in (7). As noted by K. Hunrath,364a the numbers (7) are not amicable. Nor are the ratios as given, although these ratios result if we replace 8563 by 8567 = 13-659. This false pair occurs as XIII in Euler's352 list.
(3) Problem 3 is derived from problem 2 by replacing s by a number / not necessarily prime. Let h be the greatest common divisor of ff=hg and p + l=hx. Then r+ 1 = xy, q+ 1 = gy. Also
*A11 the recnai.ni.ng cases are readily excluded. Bibliotheca Math., $), 10, 1909-10, 80-81.