234 
MR. G. T. BENNETT ON THE RESIDUES OF POWERS OF NUMBERS 
Therefore 
4+l.a+l ^a+\.a + K^\.a + 2 ^a + 2 .n + • • • + 4 + = I a +1 (mod 
'^a+2.a^\-^a+\.o. + 4 + 2 .<i +2 ^o.+2.a + • • • + %.+2.tL ^ixa = a+2 (mod 
Wll'a+l.cc + h.a+2-L-V2.a + • • • = I ^ (mod 
and we need that these shall determine a single set of values of 
L+2.a, ■ ■ ■ {mod 
Tlie necessary and sufficient condition is that the determinant 
(^(' + l,n+lj ht+2, r( + 2) • • • ^>ia) 
should be prime to p. 
Suppose that this is so, and the numbers ^a+i,a, ia+ 2 .a, • • • ^,xa determined : and 
suppose that when substituted above the values of the left-hand sides become 
-*-a+l P -L a + l> 
Next replace each modulus that exceeds p^‘ by p^% and substitute again the values 
of X. 
The first h conofruences are now satisfied. 
o 
The rest give 
(Ij+l I i + ffi (1 6 + 1 I h + \P^'' P^"'P^''{h + \.h + \ ^b + l.h 
iJ-b + 2 I 6+2P^'‘) “f" (I 6 + 2 I h+2P^'‘ ^") “ 1 “ P^'‘ (b+2.6 + 1 ^6+1.6 
+ • • • b+i.K ^p-h) = l 6 +i (mod p^) 
+ • • • '^’6+2.^ ^^6) = I5+2 (mod p'') 
(I. - iVp'") 
(iV ~ I'VP*'' +P^''(v.6+1 ^b + \.b + 
MM 
(mod 20 ^'). 
Therefore 
^6 + 1.6 + l ^6 + 1.6 d“ 4+1.6 + 2 ^6 + 2.6 “T • • • “1“ h+l./x ^ij.b = I 6 + 1 (uiod p^‘ 
>»••• ••••• •• . • . • ••••• 
G-i + l ^6 + 1.6 + 'i'i,..b + 2 ^b + 2 .b + • • • W = T^ (mod 
and we need that these shall determine a single set of values of 
^ 6 + 1 . 5 ) ^ 6 + 2 ,6> • • • (mod p'‘ ’’’). 
The necessary and sufficient condition for this is that the determinant 
(^ 6 + 1.6 + n ■^' 6 + 2 , 6+21 • • • 
should be prime to p. 
