J. B. Luce on the Theory of Numbers. 59 



For instance, let n = 216=243 2 . We have 24 = 25-1, where 

 « = 5, m = 10. These values placed in the third converging frac- 



tion, afford x=2a 2 m-3a=&85, and y=2am- 1=3 x33, and it 

 is485 2 -216x33 2 = l. 



When 



o 



to find its square factor ; to do which, we must consider that sim- 

 ple numbers are necessarily of some of the following forms, 



f? 2 ±l, £ 2 ±2, 0a-fcrf pf2±2?, <fcc. Thus if we wish to find that 



series of complex numbers which require 2 2 for their factor, I 



assume (6«±3) 2 -f 3 = 4w,(10«±5) 3 -5 = 4n,(14«±7) 2 +7=4;*, 

 (22«±ll)- + ll = 4n, &c, and making « successively 1, 2. 3, 4, 

 &c., I find the series of complex numbers 3, 21, 57, 111, 183, 

 &c., 5, 55, 155, &c, 14, 112, &c, 33, 275, &c, all of which re- 

 quire the factor 2 3 to become simple numbers. 



In like manner, for the square factor 3% I shall assume 

 (9«±4)a+2 = 9», which brings 2, 3, 19, 22, 54, 59, 107, 114, 

 178, &c. Also, 25 (9«±4) 2 + 5 = 9rc, which brings 45, 70, 470, 

 &c. all complex which are not found in the other formula. 



The formula (25 a ± 7) 2 + 1 =25n, gives 2, 13, 41, 74, 130, &c, 

 each of which complex requires the factor 5-. 



The formula (49«± 10) 2 — 2=49«, shows that the complex 



2, 31, 71, 158, &c., must be multiplied each by 7 2 . From 



(81«±22)+2=81w, we derive 6, 43, 131, 242. &c. From 

 (121«drl9) 2 +2 = 12bi, we find 3, 86, 162, &c. And other 

 similar formulas may be found for all square factors. 



If a complex be of the form a 2 -f 4, or may be brought to that 

 form by the multiplication of any square factor, we can readily 

 obtain its square factor as follows : 



5 



Let it be 29=25 + 4; which gives a = 5, m— %' Iplacethese 



values of a, and m, in the third converging fraction, and find 

 2a-°m+3a=l40, the half of which is 70: and 2<77«+l = 26, 

 whose half is 13. Wherefore we have 70* - 29 x 13 2 = - 1, or 

 70- » + l .-=29 x 13 2 ; which shows that 13 2 is the factor of 29, since 

 • 2 -f 1 i s evidently a simple number. 



Again. Let it be 61. This number is not of the form « 2 +4 ; 

 but if we multiply it by 5 2 , the product 1525=39 2 +4: which 

 worked as above, shows 3805 8 for the least square factor that 

 will make 61 a simple number. In like manner, we find that 

 109x25 is of the form « 2 + 4; hence we infer that 851325* is 

 the least square by which 109 must be multiplied to obtain a 

 simple number. 



Il is by such, and other artifices which it would be too long to 

 enumerate, that we have been able to construct a table of all the 

 square factors by which every complex number must be multi- 

 plied, to obtain a simple number, up to 158. 



