248 



Kansas Academy of Science. 



It will be evident that in a square of 4, since four numbers 

 are included in each equal sum, the sum of any line should 

 equal twice the sum of the extremes (a and q) ; and the last 

 term should equal the first term plus all the differences be- 

 tween the extremes. Written out in the form of equations 

 these statements would appear thus : 



S = 2(q + a) or S = 2g + 2a, 

 q — a = 8d + 4n + 2gr + G, and 

 g = a + 8d + 4n + 25r + G; whence, 

 S = 4a + 16d + 8w + 4sr + 2G 



A few examples will show how to use these formulae : 

 Problem. — Let it be required to write a square of 4 from a 

 varying series which will give sums of 64. 



Solution. — There are eighty-three series of integral num- 

 bers, without resorting to fractions, which will yield sums of 

 64 in a square of 4. Three such series will be selected as ex- 

 amples : 



(1) 1 2 5 6 9 10 13 14 18 19 22 23 26 27 30 31 • 



(2) 3 5 6 8 10 12 13 15 17 19 20 22 24 26 27 29 



(3) 4 5 7 8 9 10 12 13 19 20 22 23 24 25 27 28 



In the first selection it will be seen that a = 1, d = l,n = '^, 

 £^ = 3, and G = 4 ; whence 2a = 2, M = 8, An = 12, 2g = 6, 

 and G = 4. Total = 32. 



In the second selection a = S, d = 2, n = 1, g = 2, and 

 G = 2 ; whence 2a = Q, 8d = 16, 4% = 4, 2^^ = 4, and G = 2. 

 Total = 82, as before. 



In the third selection a = 4, d = 1, n = 2, g = 1, and G = 6; 

 whence 2a = S,Sd = 8, 4n = S,2g = 2, and G = 6. Total = 32, 

 half of 64. 



Putting these series into squares of 4 it will be seen that 

 they are just as perfect with these varying series as any regu- 

 lar series would make. 



(1) (2) (3) 



a=l. d=2. w=3, flr=3, G=4, a=3, d=2. m=1, g=2, G=3. o=4, d=l, n=2. g=l. G=6. 



64 64 64 



