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

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,
476 DIOPHANTUS OF ALEXANDRIA
More complicated is the case in VI. 21
2# 2 + 2# = y
2
x* + 2x 2 + x = z 3
Diophantus assumes y = mx, whence x = 2/(m 2 — 2),
/ 2 y / 2 \ 2 2
W-~2/ + Vm 2 - 2/ + m^^2 ~ *''
or 7— 5 ^
2971
= 2 3 -
(m 2 -2) 3
We have only to make 2??i 4 ,
or 2 m, a cube.
and
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 \'
II.
Method of Limits.
As Diophantus often has to find a series of numbers in
order of magnitude, and as he does not admit negative
solutions, it is often necessary for him to reject a solution
found in the usual course because it does not satisfy the
necessary conditions ; he is then obliged, in many cases, to
find solutions lying within certain limits in place of those
rejected. For example :
1. It is required to find a value of x such that some power of
it, x n ,
shall lie between two given numbers, say a and b.
Diophantus multiplies both a and b by 2 n , 3 n , and so on,
successively, until some nth power is seen which lies between
the two products. Suppose that c n lies between ap n and bp n ;
/
then we can put x = c/p, for (c
'p) n lies between a and b.
Ex. To find a square between l£ and 2. Diophantus
multiplies by a square 64; this gives 80 and 128, between
which lies 100. Therefore (V 0- 2
or ) ff solves the problem
(IV. 31 (2)).
To find a sixth power between 8 and 16.
The sixth powers
of 1, 2, 3, 4 are 1, 64, 729, 4096. Multiply 8- and 16 by 64
and we have 512 and 1024, between which 729 lies; -7 g 2 4 9 - is
therefore a solution (VI. 21).
2. Sometimes a value of x has to be found which will give