108 
PYTHAGOREAN ARITHMETIC 
(aÿoupLKoi) or recurring {olttokoltolcttcitlkol) ; these sides and 
cubes end in 1, 5, or 6, and, as the squares end in the same 
digits, the squares are called circular (kvkXlkoî). 
Oblong numbers {iTepofxrjKeLs) are, as we have seen, of the 
form m{m+1) ; prolate numbers (777)0/777/ceiy) of the form 
m (m + n) where n > 1 (c. 18). Some simple relations between 
oblong numbers, squares, and triangular numbers are given 
(cc. 19-20). If h n represents the oblong number n{n+ 1), and 
t n the triangular number ^n{n+ 1) of side n, we have, for 
example, 
h n /n 2 — {n+ 1 )/n, h n — n 2 = n, n 2 /h n _ 1 = n/{n — 1), 
n2 /hn=hn/( n + !) 2 > n 2 + (n + l) 2 + 2h n = {2n+ l) 2 , 
n 2 + h n — t^ w h n + (n +1) ^271+1’ 
n 2 ±n = J Ln 
"'n-1' 
all of which formulae are easily verified. 
Sum of series of cube numbers. 
C. 20 ends with an interesting statement about cubes. If, 
says Nicomachus, we set out the series of odd numbers 
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ... 
the first (1) is a cube, the sum of the next tiuo (3 + 5) is a 
cube, the sum of the next three (7 + 9 + 11) is a cube, and so on. 
We can prove this law by assuming that -n 3 is equal to the 
sum of n odd numbers beginning with 2 x +1 and ending 
with 2x + 2n— 1. The sum is (2x + n)w, since therefore 
(2 x + n) n = n z , 
X = £ (n 2 — n), 
and the formula is 
{n 2 — n + 1 ) + (n 2 — n + 3) + ,.. + (n 2 + 71—1) = n 3 . 
By putting successively n = 1, 2, 3 ... r, &c., in this formula 
and adding the results we find that 
l 3 + 2 3 + 3 3 + ... +r 3 = 1+ (3 + 5)+(7 + 9 +11)+... + (...r 2 + r-l). 
The number of terms in this series of odd numbers is clearly 
1+2 + 3 + ... + r or |r(r+l). 
Therefore l 3 + 2 3 + 3 3 + ... + 7’ 3 = \ r (r + 1) (1 + r 2 + r — 1) 
= {\ T (7’ + 1 ) } 2 .