478 
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;