A history of Greek mathematics - Wilbourhall.org

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508 DIOPHANTUS OF ALEXANDRIA
VI. 3,. \xy + a = u 2 .
[Suppose the required triangle to be kg, p£, bg ;
therefore
\pbg + a = a square = 7i 2 £ 2 ,
say, and the ratio of a
to n 2 -ipb must be the ratio of a square to a square.
To find n, p, b so as to satisfy this condition, form
. „ 1
a right-angled triangle trom m, — >
i.e.
therefore \pb = m 2 —
( m- + .1 m5
therefore ^2 — \pb = 4 a +
r/ (4ft 2 4-1^
or 4ft 2 + —
^
m-
??i-
4ft 2 +1
2, 771
Assume n 2 = (
1
2 ax 2
771+ )
011/
77I 2 ; and(4a + 4a 2 + 1
( m' )/
-j has to be made a square. Put
and we have a solution.
4a m 2 2 + ft (4ft 2 + 1) = (2 am + h) 2 ,
Diophantus has a =5, leading to 100m 2 + 505 = a square
= (10m + 5) 2 say, which gives m = 4 3
- and n =
; -^o --
h, p, b are thus determined in such a way that
hpb£ 2 + a = n 2 £ 2
2
g
gives a rational solution.]
VI. 4. \xy — a = u 2 .
VI. 5. a — \xy = u 2 .
VI. 6. J#2/ + # = a.
ipb£ 2 +p£ = «> an(l
[Assume the triangle to be hg, pi> r b£, so that
f°r a rational solution of this equation
we must have (ip)
2
+ a>(ipb) a square.
Diophantus
assumes p = 1, b = m, whence Ja77i + J or 2a77i+l
= a square.
But, since the triangle is rational, m 1 + 1
= a square.
That is, we have a double equation. Difference
= m 2 — 2 am — m (m — 2a). Put
2am+ 1 = {i(m — m — 2a)}
2
= a 2 , andjm = (a — 2 l)/2a.
The sides of the auxiliary triangle are thus determined
in such a way that the original equation in £ is
rationally.]
VI. 7. \xy — x — a.
solved