A history of Greek mathematics - Wilbourhall.org

500 DIOPHANTUS OF ALEXANDRIA
V. 6. x-2 = r 2 ,
y-2
= s 2 , z-2
= t
2
,
yz — y — z = u 2 , zx — z — x = v 2 , xy
— x — y = %v 2 ,
yz — x — w' 2 ,
zx — y — v' 2 , xy — z = w' 2 .
[Solved by means of the proposition numbered (3) on
p. 481.]
Lemma 1 to V. 7. xy + x 2 -\-y 2 = u 2 .
(u 2 (v-
V. 7. x 2 ±{x + y + z)= , 2 , y 2 ±(x + y + z) =
,
(v' 2
z 2 ±{x + y + z)
= 'w2
w' 2
[Solved by means of the subsidiary problem (Lemma 2)
of finding three rational right-angled triangles with
equal area. If m, n satisfy the condition in Lemma 1,
i. e. if mn + on 2 + n 2 = p
2
, the triangles are ' formed ' from
the pairs of numbers (p, m), (p, n), (p,m + n) respectively.
Diophantus assumes this, but it is easy to prove.
In his case m = 3, n £= 5, so that p = 7. Now, in
a right-angled triangle,
(hypotenuse) 2 + four times area
is a square. We equate, therefore, x + y + z to four
times the common area multiplied by £
2
, and the several
numbers x, y, z to the three hypotenuses multiplied by £,
and equate the two values. In Diophantus's case the
triangles are (40, 42, 58), (24, 70, 74) and (15, 112, 113),
and 245£ = 3360£ 2 .]
V. 8. yz±(x + y + z)=
lu 2 (v 2
j
, 2 > zx±(x+y + z) = j
,2 >
xy± (
x + y + z ) = |^2
.
[Solved by means of the same three rational rightangled
triangles found in the Lemma to V. 7, together
with the Lemma that we can solve the equations yz—a 2 ,
zx = b 2 , xy = c
2
.]
V. 9. (Cf. II. 11). x + y = 1, x + a = u 2 , y + a = v 2 .
V. 11. x + y + z= 1, x + a — u 2 , y + a=v 2 , z + a — w 2 .
[These are the problems of 7rapio-6Tr]Tos dyooyrj