A history of Greek mathematics - Wilbourhall.org

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= 2.3.7.53.4657?! = 1036648271,
X = 2. 3 2 . 89. 465771 = 7460514ft,
Y= 3 4 . 11 .465771 = 414938771,
Z- 2 2 . 5.79.465771 == 735806071,
w- 2 3 . 3. 5. 7.23. 37371= 720636071,
X = 2.3M7. 1599171 = 48932467*,
y = 3 2 . 13.4648971 = 543921371,
z = 2 2 . 3. 5. 7.11.76171 = 351582071.
It is not difficult
to find such a value of n that W+ X = a
square number; it is n = 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 i