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