INDETERMINATE EQUATIONS 
475 
a square. Where this does not hold (in IV. 18) Diophantus 
harks back and replaces the equation a? 0 —16ic 3 + a? + 64 — y 1 
by another, x ( ’ — 12 8 x 3 + x + 4096 = y 2 . 
Of expressions which have to be made cubes, we have the 
following cases. 
1. Acc 2 + Bx + C = y 3 . 
There are only two cases of this. First, in VI. 1, x 2 — 4a? + 4 
has to be made a cube, being already a square. Diophantus 
naturally makes x — 2 a cube. 
Secondly, a peculiar case occurs in VI. 17, 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 
x 3 — 3x 2 + 3x— 1 = x 2 + 2x + 3, 
or x' + x = 4 a: 2 + 4. We divide out by x 2 + 1, and x = 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 + Gx + 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 + —and where D is a cube 
o Oj 
C 
(d 3 ), y — —=x + d. Where A = a 3 and D = d 3 , we can use 
3d 2 
either assumption, or put y — ax + d. 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 = — T 2 T , 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) 
4tc + 2 = y 3 ' 
2x + 1 =z 2 
thus y 3 = 20 2 , and z — 2.