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Mathematics. — “Factorisation of large numbers” (2nd part). By 
F. J. Vars. (Communicated by Prof. P. H. Scnoure). 
V. Method of remainders. 
If G=a,? — b, (or = a? — B®) is divisible by p, then the difference 
of the remainders left by aj? and 5} after division by p must be a 
p-fold. 
If we write G= (+5 
I ie eem ey fm 
gives after 
division by p a remainder r, then it is evident that must 
give a remainder r+ 1, 
So G = (p-fold + 7 + 1)? — (p-fold + r)? = p-fold 4 2r + 1. 
If p is a factor of G, then 2r+ 1 must be a p-fold. 
Example: 
G = 80047 = (40024)? — (40023)?. 
G, — 40023 = 2002 + 23. 
So we may write 
40023 = 2002— 124+ 24 or = 201% 1994 24, 
Each of the divisors 199 or 201 will leave the remainder 24, 
As 2r+1=49 is not a 199- or 201-fold, those two numbers cannot 
be factors of G. We now find successively : 
r 2r+1 The remainders 23, 24, 27, 
Gi = 201. 199 24 49 32, ete. ascend with 1, 3, 5, 7, ete. 
202 X 198 + 27 55 Evidently 209 is a factor. 
208 X 197 + 32 65 The other factor 383 can be 
204 X 196 + 39 79 found by direct division or as 
Cla ele A canna 
206 X 194+ 59 119 a = 209 X 191 +104, so 
207X 1984+ 72 Mb GG LIP A20 HI 
208 x 192 + 87 175 — 209 x 382 + 208 4 1 
209 X 191 +104 209 Silenen, 
After two operations the factor 209 gives 11 X 19. 
With some attention the factor 11 might have been found before 
(besides by the well-known property). 
