A history of Greek mathematics - Wilbourhall.org

INDETERMINATE EQUATIONS 467
(ft) when C is positive and a square, say c 2 ;
in this case Diophantus puts Ax 2 + c 2 — (mx±c) 2 , and obtains
2mc
x =
-A +
m'
(y) When one solution is known, any number of other
solutions can be found. This is stated in the Lemma to
VI. 15. It would be true not only of the cases ±Ax 2 + C = y
2
but of the general case Ax 2 + Bx + C = y
2
. Diophantus, however,
only states it of the case Ax 2 — C — y
2
His method of finding other (greater)
.
values of x satisfying
the equation when one (x ) is known is as follows. If
A x 2 -r C = q
2
, he substitutes in the original equation (x + x)
for x and (q
— kx) for y, where k is some integer.
Then, since A (x + x) 2 — C = (q
— kx) 2 , while Ax 2 — C = q
2
it follows by subtraction that
2x(Ax + kq) — x 2 (k 2 — A),
,
,
whence
x — 2 (Ax + kq) / (k 2 — A),
and the new value of x is x -\
jo° a
'
Form Ax 2 — c
2
= y
2
.
Diophantus says (VI. 14) that a rational solution of this
case is only possible when A is the sum of two squares.
[In fact, if x = p/q satisfies the equation, and Ax 2 — c
2
= k 2 t
we have Ap 2 = c 2 q 2 + k 2 q 2 ,
Form Ax 2 + C = y
2
Diophantus proves in the
.
Lemma to VI. 12 that this equation
has an infinite number of solutions when A + C is a square,
i. e. in the particular case where x = 1 is a solution. (He does
not, however, always bear this in mind, for in III. 10 he
regards the equation 52x 2 2
+12 = y as impossible though
52 + 12 = 64 is a square, just as, in III. 11, 266a? 2 - 10 = y
2
is regarded as impossible.)
Suppose that A + C = q
2
; the equation is then solved by
Hh 2