A history of Greek mathematics - Wilbourhall.org

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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),
and
/ 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.
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