7 ° REPORT—1868. - 
observing that 24"**== —1, mod p, and transforming each of the three fac- 
torials by Sir J. Wilson’s theorem. Hence, finally, 
wal ag A ee aD Tel 
Bg Vr) gt) 9° 2 tip ida 
in accordance with the enunciation of M. Stern, The congruence A=(—1)", 
mod 4, is inferred by Eisenstein from the values of (1), W(—1), W(2), L( —2) ; 
but we may omit these determinations here. 
If p=7n+2, or 7n+4, Eisenstein considers the complex numbers formed 
with the roots of the equation n°—21n—7=0. If w is animaginary seventh 
root of unity, and 4, =3(w*+w—*)-+ 1, the roots of this equation are 7,, .5 53 
and every complex number formed with them is of the type a+6n,+en,; 
a, b, ¢ denoting real integral numbers. Let y be a primitive root of pin 
this complex theory (p is a prime of the theory, because the congruence 
n —21n—7=0, mod 9, is irresoluble: see Art. 44 of this Report), and let 
y’ =1+2,n,+4,n,, mod p, z, and z, each representing any term of the series 
0,1, 2,..p—1. The function U(w)= XY (the summation extending to all 
the p* values of ¥) is shown by Eisenstein to satisfy the equations 
Pw=Po) =o"), Yw)=V(o')=Y(w"),  Y(w) x Wo!) =p ; 
whence W(w) is of the form a+b(w+w*+o*) +¢(w°+o' +"), and 
p=A°+7B’; ifA=a—}(b+c), B=}(b—c). The equation p?>=a+3b-+3e, 
considered as a congruence, mod 7, becomes A==p*, mod 7; 7. e. A=4, 
or = 2, mod 7, according as p is of the form 7n+2 or 7n+4. To obtain 
the congruence, mod p, which is satisfied by A, we consider the congruence 
2A=V(y°)+(y*), mod p; in which e=+(p*—1)=a+(p+ yp*, a, 3, y 
representing positive integers less than p, of which the sum will be found to 
bep—1, Now W(y)=2y¥ = 3h Bh (1+2m,+e.m)*x (1 +2,n,+%.M9,)? 
x (1+2,np?+%.%,2)”, mod p,; because in general [ f(n,)?=/(np), mod p. 
Hence (y’)=0, mod p, because a+f8+y=p—l1, and_ because 
aN ae 2,91 2,%== 0, mod p, unless 9, and 6, are both different from zero, 
and both divisible by p—1. Again, if 3e=a'+/'p+y'p*, we find 
a'+'+ y'=2(p—1); and omitting terms in which the sum of the indices 
of z, and z, is inferior to 2(p—1), 
___ yP—1 5p-1 - a). ys 2 i y 
v(y*) =) Xo (2,7, +2,.) (=, + Noy) (Ngo +#,Mop2) » mod Pp: 
Substituting for = Mpr bt? Nope its value —2,9, —2,0,.—%,Np —Zn2p, We obtain 
Se, ! f—p—l ent = 1 
P(y*) =a. 116 -Iy Pe be (2:n, +2.n,)” (mp +2a%25) ’ mod p 5 
— i _ 6 —1 = . 
because every such sum as a DA " (@n, +2)" '* (mp +*.noy)” perio 
which 6 is one of the numbers 1, 2, 3,..p—1, taken positively or negatively, 
is certainly =0, mod p, as may be seen by substituting (2:mp +2009) for 
(2,n, +2,n,)?. Lastly, the coefficient C of (z,z,)?—1 in the expression oe: s+ 
(7+ zms)P "(2 Mp +2, Mop)? as evidently babi RK’, (m Noy) (n, Pg aad 
K, representing the coefficient of a in the expansion of (1+.)?~1. Hence 
C= (1y22p— Nap)? = 1, mod p, because MN2p—1s"p= £21; so that finally 
A==3 lla’. 116’. Wy' 3°" 3" (z,z,)?"' = 4 Ma’. 116. My’, mod p; 
an expression which, on substituting for a’, 6', y' their values in the two 
