A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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
which gives z — 2 n 3 — 1 + 2 ^ :i
x — n 3 = n 3 — 1, y — n 3 — it 3 ,
— 1 = 4?i 3 — 2, so that
u = ttig
— 7i (4 ii
3 — 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,
446 ALGEBRA: DIOPHANTUS OF ALEXANDRIA
when equation (1)
would give
(n — \)x = (ri z — l)u,
a solution of which is x = n 2 — 1, u = n — 1
III. The fifth problem is interesting in one respect. We are
asked to find a right-angled triangle (in rational numbers)
with area of 5 feet. We are told to multiply 5 by some
square containing 6 as a factor, e.g. 36. This makes 180,
and this is the area of the triangle (9, 40, 41). Dividing each
sicle by 6, we have the triangle required. The author, then,
is aware that the area of a right-angled triangle with sides in
whole numbers is divisible by 6. If we take the Euclidean
formula for a right-angled triangle, making the sides a . mn,
a . ^(m — 2 n
2), a . %(m 2 + n 2 ), where a is any number, and m, n
are numbers which are both odd or both even, the area is
\mn (m — n) (m + n)a
2,
and, as a matter of fact, the number mn(m — ri)(m + ri) is
divisible by 24, as was proved later (for another purpose) by
Leonardo of Pisa.
IV. The last four problems (10 to 13) are of great interest.
They are different particular cases of one problem, that of
finding a rational right-angled triangle such that the numerical
sum of its area and its perimeter is a given number. The
author's solution depends on the following formulae, where
a, b are the perpendiculars, and c the hypotenuse, of a rightangled
triangle, S its area, r the radius of the inscribed circle,
and s = %(a + b + c);
S = rs = %ab, r + s = a + b, c — s — r.
(The proof of these formulae by means of the usual figure,
namely that used by Heron to prove the formula
is easy.)
S = V{s(s-a)(s-b)(s-c)},
Solving the first two equations, in order to find a and b,
we have
a
\ = ±[r + s+ V{(r + s) 2 -8rs]],
which formula is actually used by the author for finding a