( 504 ) 
4v+2 it 
4 last figures of G, there remains ( EN 3 { OL, for the 4 last figures 
of b?, because 78 is a 4-fold + 2. 
According to the table the terminal figures 01 in a square can 
be preceded only by a 4-fold, or a 4-fold + 2, so that for b? we 
can only have: (4v + 1) 91 and for «:(40 — 1) 04. 
If we apply the same to the terminal figures 24, 44, 64, 84 of a?, 
we arrive at: 
(rije (ri), Capo) Coo 4.1) me 84 
to be subtracted 7803 
> vt2 4v—1 4v-+2 4v+2 
remainder 6 Li), GE de al, Gee 1 )S1, GEEL 
of which are only possible the cases: 
(4v 4+ 2)01, (be 1) 21, (40) 41, (40 — 1) O1, (40 4 2) 81. 
Now the first column of the additions begins with 6901, 
and therefore with a (4v-+1)01, to which is added 26060, that is 
(4 v ) 60 
which gives (4v + 1)61 and therefore never a square. 
To this is added 26260, a (4 v7 + 2) 60, together (4 v) 21, which 
can neither be a square. 
And in succession we shall have: 
(4 v ) 21 
(Av 960 
(4 v ) 81 
Ge pit from which is evident, that a square can 
(Av ) 60 never be obtained. 
(40 ) O1 So it is only necessary to calculate the 
(4 v + 2) 60 second column, which however still admits 
(40+ 2) Öl of a simplification. 
dr : = 7 En For 17401 is a (4v + 2) 01, and 26140 
(Av + 2) 60 a (4v-+1)40, so that we obtain in suc- 
40+ )81 cession : 
(Av - )60 
(Av + 2)41 
(4 v + 2) 60 
(Ev + 1) OL 
ete. 
