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

METHOD OF LIMITS 477
some function of x a value intermediate between the values
of two other functions of x.
Ex. 1. In IV. 25 a value of x is required such that 8/(x 2 + x)
shall lie between x and x + 1
One part of the condition gives 8 > x 3 + x 2 . Diophantus
accordingly assumes 8 = (o; + -§-) 3 = x a -{-x 2 + ^x + Jy,
which is
> x 3 + x 2 . Thus x + •§ = 2 or 3? = § satisfies one part of
the condition. Incidentally it satisfies the other, namely
8/(x 2 + x) < x+l. This is a piece of luck, and Diophantus
is satisfied with it, saying nothing more.
Ex. 2.
We have seen how Diophantus concludes that, if
i(a;2_60) > x > !(.£ 2 -60),
then x is not less than 1 1 and not greater than 12 (V. 30).
The problem further requires that x 2 — 60 shall be a square.
Assuming a? 2 — 60 = (x— m) 2 , we find x = (m 2 + 60)/ 2 m.
Since x > 1 1 and < 1 2, says Diophantus, it follows that
24m > m 2 + 60 > 22 m;
from which he concludes that m lies between 19 and 21.
Putting m = 20, he finds x — \\\.
478 DIOPHANTUS OF ALEXANDRIA
Diophantus assumes
whence
26+
^=( 5 +^)' or 26t/ 2 +1 = (6y + lf,
y = 10, and 1/t/ 2 = ft^
. i.e. l/x 2 = ^ft
; and
64 + ^fo = (ft)
2
.
[The assumption of 5 H— as the side is not haphazard : 5 is
chosen because it is the most suitable as giving the largest
rational value for y.']
We have now, says Diophantus, to divide 13 into two
squares each of which is as nearly as possible equal to (ft)
2
.
Now 13 = 3 2 + 2 2 [it is necessary that the original number
shall be capable of being expressed as the sum of two squares] ;
and 3 > ft by A,
while 2 < ft by ft.
But if we took 3— 5 %, 2+ ft as the sides of two squares,
2
their sum would be = 2(ft) - 5 2 2
5
-> which is
o°o
> 13.
Accordingly we assume 3 — 9#, 2 + Use as the sides of the
required squares (so that x is not exactly £§ but near it).
Thus (3-9#) 2 + (2 + lla;) 2 = 13,
III. Method of approximation to Limits.
Here we have a very distinctive method called by Diophantus
The object is to solve such
wapKroTrjs or Trapio-oTrjTos dycoyrj.
problems as that of finding two or three square numbers the
sum of which is a given number, while each of them either
approximates to one and the same number, or is subject to
limits which may be the same or different.
Two examples will best show the method.
Ex. 1. Divide 13 into two squares each of which > 6 (V. 9).
Take half of 13, i.e. 2
6J, and find what small fraction 1 /x
added to it will give a square
thus 6 J H or 26 + —
5 j >
x
y
must be a square.
and we find x = TfT
.
The sides of the required squares are ffy, f of
•
Ex. 2. Divide 10 into three squares each of which > 3
(V.ll).
xx
[The original number, here 1 0, must of course be expressible
as the sum of three squares.]
Take one-third of 10, i.e. 3 J, and find what small fraction
\/x l added to it will make a square; i.e. we have to make
1 • 9 , . 1
3 ^"l—2 a square, i.e. 30+ -j must be a square, or 30 H g
y
= a square, where 3/x = l/y.
Diophantus assumes
30i/ 2 + l = (5y + l) 2 ,
the coefficient of y, i.e. 5, being so chosen as to make 1 /y as
small as possible