286 
PROFESSOR H. J. S. SMITH ON THE ORDERS 
( 42 ) 
(1) If^ is any uneven prime, and m any given number, the congruence 
xz—y 2 =m, modp, 
admits of p solutions. 
For if ^-^^ = + 1, y 2 -\-m is prime to p for p — 2 values of y, and is divisible by p for 
2 values of y. When y 2 -\-m is prime to^?, we may assign to z any value prime to p, 
determining x by the congruence xz=y 2 -\-m; we thus obtain (p — f)(p — 2) solutions 
of (42). When y 2 -\-m is divisible by p, the congruence xz = 0, mod^?, admits of 2p — 1 
solutions; we thus obtain in all (p — f){p — 2) + 2(2p — l)=p(p-\-l) solutions of (42). 
If = — 1, y 2 -\-m is prime to p for every value of y; there are thus p(p— 1) 
solutions of (42). 
Lastly, if (yy'j = 0? i. e. if m= 0, mod p , y 2 -\-m is prime to p for p— 1 values of y, 
and divisible by p for one value of y. There are thus (p — l) 2 +2p — l=p 2 solutions 
of (42). 
We shall have to use the following corollary of this lemma, 
If m is prime to p, and if we successively attribute to x, y, z the p 3 systems of values, 
mod p , of which they are susceptible, xz — y 2 will have the same quadratic character as 
m for \p{p — 1 
of these systems. 
(2) The congruences xz—y 2 = l , 3, 5, 7, mod 8, each admit 48 solutions in which x 
and z are not simultaneously even ; of the congruences, xz—y 2 = %, =7, mod 8, the first 
admits 16, the second 48 solutions in which x and z are simultaneously even. 
For example, let the proposed congruence be xz — y 2 ^ 3, mod 8. If y has one of its 
four even values, mod 8, we may give to z any one of its four uneven values, mod 8, and 
determine the value of x in the resulting congruence ; we thus obtain 4x4 solutions in 
which x and z are uneven. If y has one of its four uneven values, mod 8, the con- 
gruence becomes xz= 4, mod 8, which admits of 8 solutions in which x and z are not 
simultaneously even, and 4 in which they are simultaneously even. There are thus 
(4x4)+(4x8)=48 solutions of the congruence [xz—y 2 = 3, mod 8, in which x and z 
are not simultaneously even, and 4x4=16 in which x and z are simultaneously even. 
(3) If is any prime, even or uneven, i and i' integral exponents, of which i> 0, i' > 0, 
and m any given number, prime to p , or divisible by any power of p , tfie congruence 
xz—y 2 =mp\ modjp i+i ' (43) 
admits of p 2i+2i ' ^1 — primitive solutions, i. e .solutions in which x, y, z, or, which is 
the same thing, x, z are not simultaneously divisible by p. 
(i) If the assertion is true for i, i', and if j <«', it is true for i+j, i'—j. For, on writing 
mjP for m in (43), it becomes 
xz—y 2 =mp i+i , mod_p (i+j0+(<W) ; 
