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

INDETERMINATE ANALYSIS 493
'
II. 34. x 2 + (x + y + z) = u 2 ,
y 2 + (x + y + z) = v 2 ,
z 2 + (x + y + z) = w 2 .
[Since {^(m — n)} 2 + mn is a square, take any number
separable into two factors (on, n) in three ways. This
gives three values, say, p, q, r for -|(m — n). Put
x = p£> y = q£, z = r£, and x + y + z = mng 2 ; therefore
(P + Q + r )i
== mng 2 ,
and £ is found.]
II. 35. x 2 — (x + y + z) = u 2 , y 2 — (x + y + z) = v 2 ,
v
z 2 — (x + y + z) = w 2 .
[Use the formula { \ (on + W-) 2 — mn — a square and
proceed similarly.]
III. 1*. (x + y + s) — x 2 = u 2 ,
(x + y + z) — y
2 = v
2
'III. 2*. (# + y + s) 2 + x = it 2 ,
(x + y + z) 2 + y =. v2 ,
III. 3*. (x + y + z) 2 -x = u 2 ,
(x + y + z) 2 -y = v 2 ,
,
(x + y + z)—z 2 = iv 2 .
(x + y + z) 2 + z = %v 2 .
(x + y + z) 2 — z = iv 2 .
III. 4*. x- (x + y + zf = u 2 ,
y-(x + y + zf = v 2 ,
— (a; + y + z) 2 = iv 2 .
III. 5. x + y + z — t
2
, y + z — x = u 2 ,
z + x — y = v 2 ,
[The first solution of this problem assumes
t 2 = x + y + z = (i + 1) 2 ,
w
2 = 1, u 2 = i
2
,
x + y — z — iv 2 .
whence x, y, z are found in terms of £, and z + x — y
is then made a square.
The alternative solution, however, is much more elegant,
and can be generalized thus.
We have to find x, y, z so that
— x + y + z = a square
x — y + z = a square
x + y — z — a square
x + y + z = a square
Equate the first three expressions to a 2 ,
b 2 , c 2 , being
squares such that their sum is also a square = k 2 ,
say.
494 DIOPHANTUS OF ALEXANDRIA
Then, since the sum of the first three expressions is
itself equal to x + y + z, we have
x = i(6f + c 2 ), y = i(c
2
+ a 2 ), z = i(«2 + & 2 ).]
III. 6. x + y + z = t 2 , y + z = u 2 , z + x = v 2 ,
x + y — w 2 .
III. 7. x — y — y — z, y + z — u 2 ,
z + x = v 2 , x + y = w 2 .
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III. 8. x + 2/ + z +