A history of Greek mathematics - Wilbourhall.org

INDETERMINATE ANALYSIS 507
(ix)
Indeterminate analysis of the fourth degree.
V. 29. x* + y* + z* = u 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 = 2 2
p
,
z 2 = q
2
2
. Therefore £4
+p* + q* = a square = (£ —
2 r) ,
say
;
2
therefore £ =
2 (r
p4 — — 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 .
(See above under V. 18.)
(x)
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 cases, in addition to the other conditions
specified.]
[Lemma to V. 7]. xy = x Y y = x Y 2 y2 .
'VI. 1. z — x = u 3 , z — y = v z .
[Form a right-angled triangle from £, m, so that
z = £2
+m 2 , x = 2m£, y — £
2 — m2
;
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 .