A history of Greek mathematics - Wilbourhall.org

HERONIAN INDETERMINATE EQUATIONS 445
number 3 is of course only an illustration, and the problem is
equivalent to the solution of the equations
(1) u + v — n(x + y))
(2) xy = n.uv )
The solution given in the text is equivalent to
x=2n 3 — 1, y — 2ti 3,
u — n(4n
3
— 2),
v — n
Zeuthen suggests that the solution may have been obtained
thus. As the problem is indeterminate, it would be natural
to start with some hypothesis, e.g. to put v = n. It would
follow from equation (1) that u is a multiple of n, say nz.
We have then
x + y = 1 + z,
while, by (2), xy — n 3 z,
whence xy = n 3 (x + y) — n 3 ,
or (x — n 3 ) (y — n 3 ) = n 3 (n 3 — 1 ).
An obvious solution is
x — n 3 = n 3 — 1, y — n 3 — it 3 ,
which gives z — 2 n 3 — 1 + 2 ^ — :i 1 = 4?i 3 — 2, so that
u = ttig
— 3 7i (4 ii — 2).
II. The second is a similar problem about two rectangles,
equivalent to the solution of the equations
(1) x + y = u + v
(2) or?/ = 7i .
m?J
and the solution given in the text is
|
x + y = u + v = n 3 — 1, (3)
u=n— 1, v = 7i (^2 — 1)^
a> = n 2 -l, 2/==7i 2 (7i~l))
In this case trial may have been made of the assumptions
v = nx, y = n 2 u,