476
DIOPHANTUS OF ALEXANDRIA
More complicated is the case in VI. 21 :
2x 2 + 2x = y 2
a; 3 + 2x 2 + x=z d ]
Diophantus assumes y = mx, whence x = 2 / (m 2 —2), and
9, m 4
or
We have only to make 2 m 4 , or 2 m, a cube.
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 bofti a and b by 2 n , 3 n , and so on,
successively, until some nih. 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 and 2. Diophantus
multiplies by a square 64; this gives 80 and 128, between
which lies 100. Therefore (f 1 ) 2 or f-| 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; is
therefore a solution (VI. 21).
2. Sometimes a value of x has to be found which will give