( 508 ) 
Av +2 
The remainder was & ig 
the lines of the 4-folds and the 4-folds — 1 on two lower lines, 
namely those on which for instance the <> of column 44 are to 
be found. 
As b? is odd, only one of those two >< can be used, namely on 
the line of the 4-folds+-2. Consequently 4” can be but a(4v+2)01, 
and therefore u° only a (4v) 04. 
The same consideration holds good for c° and d?. 
If now b? and d° were to have the same two terminal figures, then 
also d? must be a (4v + 2) 01, and c? (4v) 04. 
Now c?—a? = d’—0*; and the second member will bea (4v-+-2) 00, 
the first a (4v) 00. 
So d? and b? can never have the terminal figures alike. 
In the last example in § XIII ¢ must end in 5, so d? in 25; 
) O1 ; by the subtraction we arrive from 
: é G—1\? 
so the second and third columns will lead to (S=): 
Apparently there is an exception to the property mentioned here, 
namely in the case, that the number formed by thousands and 
4-fold 
hundreds of G are just a ree ale aie the two terminal figures 
ll 
form a number ~~ “" than the number formed by the two terminal 
greater 
figures of a?. 
For if in the example under consideration (page 505). 
Giese te 
we should have: 
es & ) of which is possible only: 
Av +3 for 4? (and so also for d*) (4v) O1 
to be subtracted 16 03 MEE Ae RE Ee 
ues 4y Then d?—-b? = (4v) 00 
semen ee + 5 se and c’—a? = (4v) 00 
Evidently the 4-folds preceding the terminal noughts must however 
be the same for both remainders, so that G would have to be 0003. 
But then we shall make use of the same consideration as above 
for the number formed by the 5 and 6" figure of G (reckoned from 
the right side). 
(March 20, 1902). 
