512 
DIOPHANTUS OF ALEXANDRIA 
so that AG = 4— Therefore (Eucl. I. 47) 
16 {n 2 -2n£ + i 2 ) = 16g 2 + 9n 2 , 
so that | = In 2 /32 n = [Dioph. has n — 1.] 
' VI. 17. ±xy + z = u 2 , x + y + z= v z . 
[Let | be the area \xy, and let z — k 2 — ^. Since 
xy = 2|, suppose x — 2, y = £. Therefore 2 + k 2 must 
be a cube. As we have seen (p. 475), Diophantus 
takes (m—l) 3 for the cube and (m+1) 2 for k 2 , giving 
m 3 — 3 m 2 + 3 m — 1 = m 2 + 2 in + 3, whence m = 4. There 
fore k = 5, and we assume \xy = z — 25 — with 
a; = 2, y = i as before. Then we have to make 
(25-|) 2 = 4 + f*, and £ = %£.] 
w VI. 18. -I«?/ + s — u 3 , a; + ?/ + z = v 2 . 
' VI. 19. \xy +■ X — u 2 , x + y + z = v z . 
[Here a right-angled triangle is formed from one odd 
number, say 2£ + l, according to the Pythagorean for 
mula m 2 + {-Km 2 — l)} 2 = {-|(m 2 + l)} 2 , where m is an 
odd number. The sides are therefore 2£+l, 2£ 2 + 2£, 
2 i 2 + 21 + 1. Since the perimeter = a cube, 
4£ 2 + 6£ + 2 = (4£ + 2) (£+ 1) = a cube. 
< Or, if we divide the sides by | + 1, 41 + 2 has to be 
made a cube. 
a • 1 2A J + 3£ 2 + £ 2£ + 1 
Again \xy + x = - = a S( l uare > 
which reduces to 2|+ 1 = a square. 
But 4£ + 2 is a cube. We therefore put 8 for the cube, 
and i = 1£.] 
VI. 20. \xy + x = U Z , X + y + Z — V 2 . 
VI. 21. X + y + z = u 2 , ^xy + {x + y + z) = v z . 
[Form a right-angled triangle from £, 1, i.e. (2 £, £ 2 —1, 
| 2 + 1). Then 2| 2 + 2£ must be a square, and £ 3 + 2 £ 2 +£