A history of Greek mathematics - Wilbourhall.org

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470 DIOPHANTUS OF ALEXANDRIA
In order that this equation may reduce to a simple equation,
either
(1) the coefficient of x 2 must vanish, or oc — /? = 0,
or (2) the absolute term must vanish, that is,
or
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p 2 (a + b) + (a- b)
2
= 0,
{ p 2 — (a + b) }
2 = 4 ab t
so that ab must be a square number.
As regards condition (1) we observe that it is really sufficient
2 is equally
if ocn 2 = /3m 2 , since, if oc x -f a is a square, (otx + a)
a square, and, if fix + b is a square, so is (ftx + tym 2 , and
vice versa.
That is, (1) we can solve any pair of equations of the form
ocm 2 x + a = w 2]
an 2 x + b — w 2
Multiply by n 2 , m
equations
2 respectively, and we have to solve the
ocm 2 n 2 x+an 2 = u' 2
oc m 2 n 2 x + bm 2 = w' 2 \
Separate the difference, an 2 — bm 2 ,
put u f ± iv' = p,
into two factors p, q and
u'+w' = q\
therefore u' 2 2
= \(p + q)
, w' 2 = \(p — q)
2
,
and a m 2 n 2 x + an 2 = %(p +