A history of Greek mathematics - Wilbourhall.org

INDETERMINATE ANALYSIS 505
Diophantus assumes u = £ + 1 , v = £ — 1 , whence
i(6f+2) 2 -2(2£* + 6£)
must be a square, or
9£ 4 -4£ 3 2
+ 6£ -12£+l:=a square = (3 £ 2 -6£+l) 2 , say;
therefore 32 £
3 = 36 £
2
, and
£ = §. Thus u, v are found,
and then x, y.
The second (alternative) solution uses the formula that
§(i*-i) + (i*
T £) + i = a cube. Put x = $, y= f -|,
and one condition is satisfied. We then only have to
make £(f2 -£) -£- 2 3
(£ -£) or -2£ £ 2
a>i.e.£ 3 -2£ 2 =(i^say.]
a cube (less than
IV. 38. (x + y + z)x = ^u(u + 1), (x + y + z)y — v 2 ,
[Suppose # + 2/ + s = £
2
(x + y + z)z = w 3 ,
[x + y + z = t 2 ~\.
;
then
_ 16 (u + 1 ) v 2 u> 3