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

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INDETERMINATE EQUATIONS 469
470 DIOPHANTUS OF ALEXANDRIA
1 . Double equation of the first degree.
The equations are
either
a x + a = u 2 ,
+ b — w 2 .
or (2) the absolute term must vanish, that is,
Diophantus has one general method taking slightly different
forms according to the nature of the coefficients.
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p 2 2
(a + b) + (a- b)
= 0,
or
(a) First method of solution.
{ p — 2 (a
2
+ b) } = 4 ab t
This depends upon the identity
so that ab must be a square number.
{i(p+q)V'-{i(p-q)V- = pqif
ocn 2 = /3m 2 , since, if oc x -f a is a square, (otx + a)
If the difference between the two expressions in x can be
separated into two factors p, q, the expressions themselves
are equated to {i(p +
2
q}} and { \ (p — 2 q) }
respectively. As
'
Diophantus himself says in II. 11, we equate either the square
of half the difference of the two factors to the lesser of the
ocm 2 x + a = w 2]
an 2 x + b — w 2
Multiply by n 2 , m
Diophantus's point of view, remembering
equations
that the final quadratic
in x must always reduce to a n 2 x+an 2 = u' 2
ocm
single equation.
oc m
Subtracting, we have (oc — ft) x + (a — b) = u — n 2 x + bm = 2 w' 2 \
2
w 2 .
Separate the difference, an
Separate (oc — ft) x + (a — — bm 2 ,
b) into the factors
put u f ± iv' = p,
p, {(a-ft)x + (a-b)} /p.
u'+w' =
We q\
write accordingly
(oi—ft)x + (a-b)
therefore u' =
n + w — — \(p +
2
q)
, w' = 2 \(p —
-,
P
and a m 2 n 2 x + an 2 2
= %(p +