A history of Greek mathematics - Wilbourhall.org

;
—
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,
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
y g
= 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