A history of Greek mathematics - Wilbourhall.org

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472 DIOPHANTUS OF ALEXANDRIA
or, since pq = (xd 2 — @c 2 ,
p 2 x 2 -2x((xd 2 + l3c 2 ) + q
2
-4e 2 d 2 = 0.
In order that this may reduce to a simple equation, as
Diophantus requires, the absolute term must vanish, so that
The method therefore only gives one solution, since
q = 2cd.
q is restricted to the value 2cd.
Ex. from Diophantus
Sx + 4 =u 2 )
6^ + 4 = w 2 )
/TTr
(IV. 39)
Difference .2 x; q necessarily taken to be 2 a/4 or 4; factors
therefore \x, 4. Therefore Sx + 4 = \ {\x + 4) 2 , and a; = 112.
(/?) Second method of solution of a double equation of the
first degree.
There is only one case of this in Diophantus, the equations
being of the form
hx + n 2 = u 2 )
(h+f)x + n 2 =w 2 )
Suppose hx + n 2 = (y + n) 2 ;
therefore hx = y
2
+ 2 ny,
•f
and (h +f ) x + n 2 — (y + n) 2 + r (y 2 + 2 ny).
It only remains to make the latter expression a square,
which is done by equating it to {jiy — nf.
The case in Diophantus is the same as that last mentioned
(IV. 39). Where I have used y, Diophantus as usual contrives
to use his one unknown a second time.
2. Double equations of the second degree.
The general form is
Ax 2 +Bx +C = u 2 \
A'x 2 + B'x + C'=iv 2 \
'
but only three types appear in Diophantus, namely
. .
p
2 x 2 + (xx + a — u
2
(1) J-
p 2 x 2 + px + b — w l )
) .
, where, except in one case, a = 6.
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