472 
DIOPHANTUS OF ALEXANDRIA 
or, since 
pq = ad 2 —(3 c 2 , 
p 2 x 2 — 2 x{pi.cl 2 + (3c 2 ) + q 2 — 4 c 2 d 2 = 0. 
In order that this may reduce to a simple equation, as 
Diophantus requires, the absolute term must vanish, so that 
q = 2cd. The method therefore only gives one solution, since 
q is restricted to the value 2 cd. 
Ex. from Diophantus: 
8x + 4 = u 2 ' 
6x + 4 = w 2 
(IV. 39) 
Difference 2x; q necessarily taken to be 2^4 or 4; factors 
therefore ^x, 4. Therefore 8cc + 4 = (^¡r + 4) 2 , and x — 112. 
(/3) 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 
= u 2 1 
{h +f)x + n 2 = w 2 j 
Suppose hx + n 2 = {y + n) 2 : therefore hx = y 2 + 2 ny, 
and 
It only remains to make the latter expression a square, 
which is done by equating it to {py — n) 2 . 
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 
but only three types appear in Diophantus, naraely 
^ ^ p 2 x 2 + (3x + b = w 2 
p 2 x 2 + ax + a = u 2 
where, except in one case, a = b.