510 
DIOPHANTUS OF ALEXANDRIA 
Lemma 1. If a rational right-angled triangle be formed 
from on, n, the perpendicular sides are 2 mn, on 2 — n 2 . 
We will suppose the greater of the two to be 2 mn. 
The first two relations are satisfied by making m = 2 n. 
Form, therefore, a triangle from £, 2£. The third con 
dition then gives 6 £ 4 + 3 £ 2 = a square or 6 £ 2 + 3 =. a 
square. One solution is £ = 1 (and there are an infinite 
number of others to be found by means of it). If | = 1, 
the triangle is formed from 1,2. 
VI. 12. Suppose the triangle to be (/¿¿, h p £). Then 
iiP b )ê 2 +Pê = a square = (/vI) 2 , say, and i=p/(h 2 —ipb). 
This value must he such as to make {^pb)£ 2 + a square 
also. By substitution of the value of ^ we get 
{bpk 2 + ^p 2 b{po — b)]/{k 2 — \pb) 2 ; 
so that bpk 2 + ^p 2 b(p — h) must be a square; or, if p, 
the greater perpendicular, is made a square number, 
bk 2 + ^pb{p — b) has to be made a square. This by 
Lemma 2 (see p. 467 above) can be made a square if 
h + \ pb ip — b) is a square. How to solve these problems, 
says Diophantus, is shoivn in the Lemmas. It is not 
clear how they were applied, hut, in fact, his solution 
is such as to make p, p — b, and b + ^pb all squares, 
namely b = 3, p = 4, h = 5. 
Accordingly, putting for the original triangle 3|, 4£, 5|, 
we have 
6 £ 2 + 4 i — a square ) 
6+ 3 £ = a square j 
Assuming 6£ 2 + 41 = m 2 | 2 , we have £ = 4/{on 2 — 6), and 
the second condition gives 
96 
12 
or 
This can be solved, since m = 1 satisfies it (Lemma 2). 
A solution is on 2 = 25, whence | = T 4 g.] 
| VI. 14. \xy — z = oi 2 , \xy — x = v 2 .