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DIOPHANTUS OF ALEXANDRIA
Diophantus assumes
26 + p = ( 5 + “• 26 2/ 2 + 1 = (5y+ l) 2 ,
whence
у 10, and 1 / — тпо; be. 1/ж 2== ?5o) and 6^- + 4qq = (-|^-)-,
[The assumption of 5 + ^ 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 (-|£) € .
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 о >• jo Py 2 0 >
while 2 < by
But if we took 3 —2 9 q, 2+i^- as the sides of two squares,
their sum would be 2(-|£) 2 = which is > 13.
Accordingly we assume 3 — 9ж, 2 + llccas the sides of the
required squares (so that x is not exactly gV but near it).
Thus (3 — 9x) 2 + (2 4-11 x) 2 = 13,
and we find x — T f T .
The sides of the required squares are ffy, fff.
Ex. 2. Divide 10 into three squares each of which > 3
(V. 11).
[The original number, here 10, must of course be expressible
as the sum of three squares.]
Take one-third of 10, i.e. 3|, and find what small fraction
1 /x 1 added to it will make a square; i.e. we have to make
19 1
3-j-)—- a square, i.e. 30+ must be a square, or 30-1—
x x^ у
= a square, where 3 /x — 1 /у.
Diophantus assumes
30^+1 ={5y+ l) 2 ,
the coefficient of y, i.e. 5, being so chosen as to make 1 /у as
small as possible;