A history of Greek mathematics - Wilbourhall.org

INDETERMINATE EQUATIONS 475
a square. Where this does not hold (in IV. 18) Diophantus
harks back and replaces the equation x G 2
— I6x + ?j 02 + 64 = y
by another, a? — 6 128a; 3 2
+ x + 4096 = y
.
Of expressions which have to be made cubes, we have the
following cases.
1. Ax 2 + Bx + C = y
3
.
There are only two cases of this. First, in VI. 1, 032 -— 4 a; + 4
has to be made a cube, being already a square. Diophantus
naturally makes «-2a cube.
Secondly, a peculiar case occurs in VI. 1 7, where a cube has
to be found exceeding a square by 2. Diophantus assumes
(x— l) 3 for the cube and (x + l) 2 for the square. This gives
a; 3 -3a; 2 +3a:-l = x 2 + 2x + 3,
or x"' + x — 4 a?2 + 4. We divide out by x 2 +l, and a; = 4. It
seems evident that the assumptions were made with knowledge
and intention. That is, Diophantus knew of the solution 27
and 25 and deliberately led up to it. It is unlikely that he was
aware of the fact, observed by Fermat, that 27 and 25 are the
only integral numbers satisfying the condition.
2. Ax 3 + Bx 2 + Cx + D*= y
3
, where either A or D is a cube
number, or both are cube numbers. Where A is a cube (a 3 ),
we have only to assume y = ax+ —-r } ,
G
(d 3 ), y— —-j 9
x + d. Where A = a 3 and D = d 3 , we
o CL"
either assumption, or put y — ax + d.
and where D is a cube
can use
Apparently Diophantus
used the last assumption only in this case, for in IV. 27 he
rejects as impossible the equation 8x 3 — x 2 + 8x—l=y 3 ,
because the assumption y = 2x— 1 gives a negative value
x = — xx, whereas either of the above assumptions gives
a rational
value.
(2) Double equations.
Here one expression has to be made a square and another
a cube. The cases are mostly very simple, e.g. (VI. 19)
thus y 3 — 2z 2 ,
and z = 2.
4a; + 2 = y
3 )
2x + \
=Z 2 \'