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

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
whence
2x(Ax + kq) — x 2 (k 2 — A),
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 ,
'
,
,
468 DIOPHANTUS OF ALEXANDRIA
substituting in the original equation 1 + x for x and (q
— kx)
for y, where k is some integer.
3. Form Ax 2 + Bx + C=y 2 .
This can be reduced to the form in which the second term is
wanting by replacing x by z — —j
•
2 XL
Diophantus, however, treats this case separately and less
fully. According to him, a rational solution of the equation
Ax' 1 + Bx +C = y
2
is only possible
(a) when A is positive and a square, say a 2 ;
(/?) when C is positive and a square, say c 2 ;
(y)
when ^B 2 — AC is positive and a square.
In case (a) y is put equal to (ax — m), and in case (/3) y is put
equal to (mx — c).
Case (y) is not expressly enunciated, but occurs, as it
were, accidentally (IV. 31). The equation to be solved is
3 x + 1 8 — x 2 = y
2
. Diophantus first assumes 3 x + 1 8 — x 2 = 4 x 2 ,
which gives the quadratic 3^+18 = 5 sc 2 ;
but this 'is not
rational '. Therefore the assumption of 4 x 2 for y 2 will not do,
'
and we must find a square [to replace 4] such that 1 8 times
(this square + 1 ) 4- (f )
2
may be a square '. The auxiliary
equation is therefore 18(m 2 2
+ 1) + § = y
, or 7 2 m 2 + 81= a
square, and Diophantus assumes 72 m 2 + 8 1 = (8 m + 9)
2
, whence
m= 18. Then, assuming 3 x + 18 — x 2 = (1 8) 2 ^ 2 , he obtains the
equation 325 2 — 3x— 18 = 0, whence x — /2t> ^na^ is > 2V
(2) Double equation.
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
The Greek term is SLTrXoicroT-qs, SlttXtj IcroT'qs or SlttXyj io-cdctis.
Two different functions of the unknown have to be made
simultaneously squares. The general case is to solve in
rational
numbers the equations
mx 2 + oc x + a — u 2 j
nx 2 -\- fix+ b = w 2 )
The necessary preliminary condition is
that each of the two
expressions can be made a square. This is always possible
when the first term (in x 2 ) is wanting. We take this simplest
case first.