CHAP. V] EULEK'S ^-FUNCTION. 137
G. A. Miller101 proved (4) by noting that in a cyclic group G of order N there is a single cyclic subgroup of order df a divisor of N, and it contains <t>(d) operators of order d, while the order of any operator of G is a divisor of N. Thus (4) states-merely that the order of G equals the sum of the numbers of the operators of the various possible orders. Next, (1) follows from an enumeration of the operators of highest period A B in a cyclic group of order AB, which is the direct product of its cyclic subgroups of orders A and B. Finally, if p is a prime, all the subgroups of a cyclic group of order pn are contained in its subgroup of order pn~l, whence <£(pn) = pn —pn~l.
G. A. Miller102 proved the last three theorems and the fact that </>(Z) is even if l>2 by means of the properties of the abelian group whose elements are the integers < m which have with m a g. c. d. equal to k.
"K. P. Nordlund103 proved ct>(mn...) = (m — 1)(n — 1)..., where m,n,... are distinct primes, by writing down the multiples <mnp of m, the multiples of ran, etc., whence the number of integers <mnp and not prime to it is mnp — 1 — (m — 1) (n—1) (p — 1).
E. Busche104 treated geometrically systems (JM) of four integers such that ad—6c>0, evaluated the number $(/S) of systems incongruent modulo S and prime to S, and generalized (4) to S$(S).
L. Orlando105 showed that 4>(m) is determined by (4) [Lucas72].
H. Bonse106 proved Maillet's93 theorem for r = l, 2, 3 without using Tchebychef s theorem. His lemma was generalized by T. Suzuki.106a
J. Sommer107 gave without reference Crelle's8 final evaluation of <f>(ri).
R. D. Carmichael108 proved that if n is such that <£(aO=n is solvable there are at least two solutions x. He found solutions of <t>(x) = 2n [in accord with Pichler94] and proved that there are just ti+2 solutions (a single one being odd) when n^31 and just 33 solutions when 32^n^255. All the solutions of <fr(x) =4n—2> 2 are of the form pa, 2pa, where p is a prime of the form 4s —1; for example, if n = 5, the solutions are 19, 27 and their doubles.
Carmichael109 gave a table showing every value of m for which <£(w) has any given value g 1000.
A. Ranum109a would solve <t>(x)~n by resolving n in every possible way into factors n0,., nr, capable of being taken as the values of <K2ao), 4>(piai), . .., <Kpr°r)> where 2, pi,. .., pr are distinct primes. Then 2aop1ai.. .prar is a value of x.
Carmichael110 gave a method of solving </>(#)= a, based on the testing of the equation for each factor x of a definite function of a.
M. Fekete111 considered the determinant pkn obtained by deleting the last row and last column of Sylvester's eliminant for xk — 1 = 0 and xn — 1 =0
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