Chap. I] PERFECT, MULTIPLY PERFECT, AND AMICABLE NUMBERS. 43 
not all prime. In the next case w = 6, /=257, Euler examined only the 
case* x — 6 = 2^-257, finding q composite. 
For A; = 2, Euler excluded m = 1, 3 [m = 4 is easily excluded]. 
(13) For k^n in (I2), = 2*", where m = k—n. Then 
6 = 2"+i+2"'+"-l=/ 
must be a prime. Thus we must take as the factors of 6^ 
2'"x-6 = l, 2"'2/-fe = 62, 
whence z = 2" +2"+^"'", y-=hx. If m = 1, one of 
/=2'»+2-l, p = 2^+'-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 = 2"(gr— l)(/i — 1), where the last two factors are prime, set 
d = 2a-fa. Then 
ig - 2^*+^) {h - 2^^+^) =d- 2"+^ + 2^""+^ 
Euler treated the cases n^3, d = 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 = 3^.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 jP'j2 
— ['>''] s, we may set 
p = ax — l, q = ^y — l, r = /3x — 1, s = ay — l. 
We set fa:a = 2b—c:h, where b and c are relatively prime. The second 
condition fa- fpq = a{pq+rs) gives 
ca/3x2/ = 6(a+iS) (x+2/) -26. 
Multiply it by ca^. Then 
[ca^x-h{a+^)][ca^y-b(a+^)] = h\a+^y-2hca^. 
Given a, ^, a and hence b, c, we are to express the second member as a prod- 
uct of two factors and then find x, y. 
For a = l, i3 = 3, = 2^ Euler obtained the pairs (a), (28). For a = 2, 
^ = 3, a = 32-5-13, he got (32); for a = l, /3 = 4, a = 3^-5, (30). The ratio a:^ 
may be more complex, as 5 :21 or 1 :102, in (7). As noted by K. Hunrath,^^^* 
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's^^^ 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 
andp+l = ^x. Thenr-{-l=xy, q-{-l = gy. Also 
ghxyfa=f{afr)=a(pq+fr)^a\{hx-l){gy-l)+f{xy-l)\. 
*A11 the remaining cases are readily excluded. 
"♦"BibUotheca Math., (3), 10, 1909-10, 80-81. 
