270 MR. G. T. BENI^ETT OR' THE RESIDUES OF POWERS OF RUMBERS 
If X is even and = 2ja, then we may have 5 = />(, + 1, and in this case X — 25 = 2. 
We wish, therefore, to find the numbers with ex|)onent 2'^“^ for mod (1 + 
We have to take the solutions of 
^ = 1 [mod (1 + 
and exclude those of 
^ = 1 [mod (1 + 
The first leads to 
0 / = d: 1 [mod (1 + ^)*], 
and the second to 
a = d: 1, d: * [mod (1 + 7)^], 
therefore 
CL — 1 -J~ 2 ' 2 ', 2 d" 3 d~ 2?, 2 -]~ 3^ [mod (1 -j~ 
Hence the numbers with exponent 2'"“^ for mod (1 + are 
1 d" 2'?', 2 d" h 3 d“ 2i, 2 Si [mod [1 d~ 
The number of them is 4 X N (1 + 
— 2h 2^^*“^ 
For mod (1 + there are 
2^'^ ^ numbers with 
exp 2^* 
22 m -3 _p 
J? 
„ 2-^ 
22^ + 1 _p 22 ^ 
5 ? 
„ 2^ 
•7 7 1 ufi 
W ^ JJ 
? > 
„ 23 
2" d- 2“^ d- 23 
J? 
» 23 
2- d- 2 d- 1 
? 5 
„ 2 
1 1 ' 
)) )» )j I J 
making, in all, 2"'*“^ = <I> (1 + 7)"'^ numbers as it should. 
(xvi.) Generators for the modulus (1 + ^)^ 
If X = 2/r + 1, it is easy to see that in order to generate 2“'" numbers, of which for 
each exponent tliere shall be the right number of numbers (as just found), we must 
take three generators :— 
two generators with exp 2'^ “ H 
and p • 
one genei'ator with exp 2^ J 
