INDETERMINATE ANALYSIS 507 
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(ix) Indeterminate analysis of the fourth degree. 
V. 29. x i + y i + z i = 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 = p 2 , 
z 2 = q 2 . Therefore £ 4 +p i + q i — n square = (£'~ — r) 2 , say; 
therefore | 2 = {r 2 ~p i — q i )/2r, and we have to make 
this expression a square. 
Diophantus puts r = p 1 + 4, f — 4, so that the expres 
sion reduces to 8p 2 /(2p l + 8) or 4p 2 /{p 2 + 4). To make 
this a square, let p 2 + 4 = (p + l) 2 , say; therefore p— \\, 
and p 2 =2%, rf — 4, r = 6|q or (multiplying by 4) 
p 2 = 9, q 2 = 16, r = 25, which solves the problem.] 
[V. 18]. x 2 + y 2 + z 2 — 3 = u 4 . 
(See above under Y. 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 0, 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 x y x = xpj 2 . 
'VI. 1. z — x — u 3 , z — y = v 3 . 
[Form a right%ngled triangle from m, so that 
s = £ 2 + m 2 , x = 2 m£, y = £ l — m 2 \ thus z — y= 2 m 2 , 
{ and, as this must be a cube, we put m = 2; therefore 
z — x = £ 2 — 4£ + 4 must be a cube, or £ — 2 = a cube, 
say n 3 , and | = n 3 + 2.] 
(VT. 2. z + x = u 3 , z + y = v 3 .