A history of Greek mathematics - Wilbourhall.org

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£
512 DIOPHANTUS OF ALEXANDRIA
so that AG = 4 (n
£). Therefore (Eucl. I. 47)
16(n — 2 2
2n£ + g )
= 16£
2
+ 9^ 2
,
so that £ = 7 ri 1 / 32 n = ^w. [Dioph. has n = 1.]
r
VI. 17. \xy-\-z = ti 2 ,
cc + 2/ + ^ = u 3 .
[Let £ be the area Jar?/, and let z — k 2 — £. Since
xy = 2£, suppose x = 2, y = £. Therefore 2 + & 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 — 3m 2 + 3 m — 1 =m 2 + 2 m + 3, whence rn = 4. Therefore
A; = 5, and we assume \xy = £ s = 25 —
3 £, with
a? = 2, y = i as before. Then we have to make
(25-£) 2 = 4 + £
2
, and
£ = «?-.]
VI. 1 8. -!#?/ + = u 3 , x + y ±z = v 2 .
"
vi. i9. 4^2/ + x — u2 >
®+y+z
= v 3 '
[Here a right-angled triangle is
number, say 2£+l, according to
formed from one odd
the Pythagorean formula
m 2 + {-|(m 2 — l)} 2 = {-|(m 2 +l)} 2 ,
where m is an
odd number. The sides are therefore 2£+l, 2£ 2 +2|,
2 £
2
-f 2 £ + 1 . Since the perimeter == a cube,
4£ 2 + 6£ + 2 = (4£ + 2) (£+1) = a cube.
Or, if we divide the sides by £+1, 4 £ + 2 has to be.
made a cube.
• 1
a
2£ 3 + 3£ 2 + £ 2£+l
Again \xy + x = *
/