98 
ARCHIMEDES 
Secondly, it is required that 
W+ X = a square, (6) 
Y+Z = a triangular number. (i) 
There is an ambiguity in the text which makes it just possible 
that W+X need only be the product of two whole numbers 
instead of a square as in (0). Jul, Fr. Wurm solved the problem 
in the simpler form to which this change reduces it. The 
complete problem is discussed and partly solved by Amthor. 1 
The general solution of the first seven equations is 
W= 2.3.7.53.4657n= 1036648277, 
X= 2.3 2 .89.4657n = 746051471, 
Y = 3 4 .11,4657% = 414938771, 
Z= 2 2 .5.79.465777 = 735806077, 
w — 2 3 .3.5.7.2 3.3 7 3 77 = 720636077, 
x = 2.3 2 .17 . 1599177 = 489324671, 
y = 3 2 .13.4648977 = 543921377, 
Z — 2 2 .3.5.7.11.76177= 351582077. 
It is not difficult to find such a value of n that W+X = a 
square number; it is 77 = 3 . 11.29.4657£ 2 = 4456749£ 2 , 
where £ is any integer. We then have to make Y + Z 
a triangular number, i.e. a number of the form •§<?(#+l)- 
This reduces itself to the solution of the ‘ Pellian ’ equation 
t 2 ~ 4729494U 2 = 1, 
which leads to prodigious figures; one of the eight unknown 
quantities alone would have more than 206,500 digits! 
(ß) On semi-regular polyhedra. 
In addition, Archimedes investigated polyhedra of a certain 
type. This we learn from Pappus. 2 The polyhedra in question 
are semi-regular, being contained by equilateral and equi- 
1 Zeitschrift für Math. u. Physik (Hist.-litt. Abt.) xxv. (1880), pp. 
156 sqq. 
2 Pappus, v, pp. 352-8.