264 HISTORY OF THE THEORY OF NUMBERS. [CHAP, rx H. D. Carmichael10 treated the problem to find m, given the prime p and s =Soi, in Legendre's formula; a given solution m2 leads to an infinitude of solutions map*, k arbitrary. Again, to find m such that pm-< is the highest power of p>2 dividing ml, we have m— t = (m — s)/(p — 1), and see that w has a limited number of values; there is always at least one solution m. Carmichael11 used the notation H\y\ for the index of the highest power of the prime p dividing y, and evaluated U=o where a, c are relatively prime positive integers. Set c0=c and let v be the least integer such that ia+Cr-i is divisible by p, the quotient being cr. Let Then /i=2(er+l), where iis the least subscript for which r-l c,(a+c,)(2a+c<) . ..( is not divisible by p. It follows that where E is the index of the highest power of p not exceeding n — 1 . If n is a power of p, /*= (n-l)/(p-l). But if n = 5kpk+ . . . +5^+^, 5A^0, and at least one further 8 is not zero, . . . . p-l p~l In case the first x for which xa+c is divisible by p gives c as the quotient, all the cr are equal and hence all the tr; then , - [n — l-i+P L p The case a = c = l yields Legendre's2 result. The case a = 2, c = l, gi gves E. Stridsberg12 wrote JYm for (1) and considered where a is any integer not divisible by the positive integer m. Let p be a prime not dividing m. Write a,- for the residue of aj modulo m. He noted that, if pj=l (mod m), 10Bull. Amer. Math. Soc., 14, 1907-8, 74-77; Arncr. Math. Monthly, 15, 1908, 15-17. »/Wd.f 15, 1908-9, 217. lxArkiv for Matematik, Astr., Fyaik, 6, 1911, No. 34.