A history of Greek mathematics - Wilbourhall.org

m
INDETERMINATE ANALYSIS 499
IV. 29. x 2 +
2
y + z 2 + iv 2 + x + y + z + w = a.
[Since x 2 + x+% is a square,
(x 2 + x) +
2
(y + y) + (z 2 + z) + (w 2 + iv) + 1
is the sum of four squares, and we only have to separate
a + 1
into four squares.]
I IV. 30. x 2 + y
2
+ z 2 + iu 2 — (x + y + z + w) = a.
IV. 31. x + y — 1, (# + a) (# + &) = w 2 .
IV. 32. ^ + 2/4-^ = 0-, xy + z = u 2 , xy — z = v 2 .
IV. 39. x — y = m(y — z), y + z = u 2 , z + x = v 2 , x + y = to 2 .
IV. 40. x 2 — y
2
= m(y — z), y + z = u 2 , z + x = v 2 , x + y = w 2 .
V. 1. xz = y
2
, x
— a = u 2 , y
— a = v 2 , z — a = w 2 .
V. 2. xz — y
2
, x
+ a =u 2 , y
+ a = v 2 , z+a
— w 2 .
( V. 3. x + a = r 2 , y + a = s 2 , z + a = £ 2 ,
2/0 + (X = u 2 , ;sa? + a = v
2, xy + a = iy 2 .
V. 4. a? — a = r 2 , y — a = s 2 , z — a — t 2 ,
yz — a— u 2 , zx
— a=v 2 , xy
— a = w 2 .
[Solved by means of the Porisms that, if a be the
given number, the numbers m 2 — a, (m+1) 2 — a satisfy
the conditions of V. 3, and the numbers m 2 + a,
(m + l) 2 + a the conditions of V. 4 (see p. 479 above). The
third number is taken to be 2 {m 2 + a + (m + l) 2 + a} — 1,
and the three numbers automatically satisfy two more
conditions (see p. 480 above). It only remains to make
2 {m 2 + a + (m + 1) 2 + a] — 1 +a & square,
or 4
2
+ 4m + 3 a + 1 = a square,
which is easily solved.
With Diophantus £ + 3 takes the place of m in V. 3
and £ takes its place in V. 4, while a is 5 in V. 3 and 6
in V. 4.]
V. 5.
y 2 z 2 + x 2 = r 2 , z 2 x 2 + y
2
= s 2 , x
2 y 2 + z 2 = t 2 ,
y 2 z 2 +
2
y + z — 2 u 2 , z 2 x 2 + z 2 + x — 2 v 2 , x 2 y 2 + x 2 2
+ y = w 2
[Solved by means of the Porism numbered 2 on p. 480.
K k 2