A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

INDETERMINATE ANALYSIS 507
508 DIOPHANTUS OF ALEXANDRIA
(ix)
Indeterminate analysis of the fourth degree.
V. 29. x* + y* + z* = u 2 .
2 2
[' Why ', says Fermat, did not Diophantus seek two
'
fourth powers such that their sum is a square. This
problem is, in fact, impossible, as by my method I am
able to prove with all rigour.' No doubt Diophantus
= knew this truth empirically. Let x 2 2
= £
,
y p
,
z 2 = q
2
2
. Therefore £4
+p* + q* = a square = (£ —
2 r) ,
say
;
2
therefore £ =
2 (r — 4 — g 4 )/2r, and we have to make
this expression a square.
Diophantus puts r = £>2 + 4, g
2 = 4, so that the expression
reduces to 8p 2 /(2p 2 + 8) or 4p 2 /(p 2 + 4).
To make
this a square, let
p 2 + 4 = (p + l) 2 , say ; therefore p = 1-|,
and p 2 =2%, q
2
= 4, r = 6^; or (multiplying by 4)
£> 2 =9, q
2
= 16, r = 25, which solves the problem.]
[V. 18]. ^2 + 2/
2
+ 2 2 -3 =u 4 .
(x)
(See above under V. 18.)
Problems of constructing right-angled triangles with
sides in rational numbers and satisfying various
other conditions.
[I shall in all cases call the hypotenuse z, and the
other two sides x, y, so that the condition x 2 + y
2
= z 2
applies in all
specified.]
[Lemma to V. 7]. xy = x Y y Y = x 2 y2 .
'VI. 1. z — x = u 3 , z — y = v z .
cases, in addition to the other conditions
[Form a right-angled triangle from £, m, so that
z = £2
+m 2 , x = 2m£, y — £
2 — 2 ;
thus z — y = 2m 2 ,
and, as this must be a cube, we put m = 2 ;
therefore
2 — # = £
2
— 4£ + 4 must be a cube, or £ — 2 = a cube,
say % 3 , and £ = ?i 3 + 2.]
VI. 2. s + & = u 3 , z + y = v 3 .
VI. 3,. \xy + a = u 2 .
[Suppose the required triangle to be kg, p£, bg there-
;
say, and the ratio of a
fore \pbg + a = a square = 7i 2 £ 2 ,
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 m
5
therefore ^2 — \pb = 4 a +
??i-
4ft 2 +1
2, 771
Assume n 2 = (
1
2 ax 2
771+ )
011/
77I 2 ; and(4a + 4a 2 + 1
( m' )/
r/ (4ft 2 4-1^
or 4ft 2 + —
^ -j has to be made a square. Put
m-
4a m 2 2 + ft (4ft 2 + 1) = (2 am + h) 2 , and we have a solution.
Diophantus has a =5, leading to 100m 2 + 505 = a square
= (10m + 5) 2 say, which gives m = 2 3 -
; g4
and n = -^o --
h, p, b are thus determined in such a way that
hpb£ 2 + a = n 2 £ 2
VI. 4. \xy — a = u 2 .
VI. 5. a — \xy = u 2 .
VI. 6. J#2/ + # = a.
gives a rational solution.]
[Assume the triangle to be hg, pi> r b£, so that
ipb£ 2 +p£ = «> an(l f°r a rational solution of this equation
we must have 2
(ip) + 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