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

INDETERMINATE ANALYSIS 499
IV. 29. x 2 + y
2
+ z 2 + iv 2 + x + y + z + w = a.
[Since x 2 + x+% is a square,
(x 2 + x) + (y
2
+ 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 .
2
V. 1. xz = y
, x
2
V. 2. xz — y
, x
— a = u 2 , y
+ a =u 2 , y
— a = v 2 , z — a = w 2 .
+ a = v 2 , z+a
( 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 .
— 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,
500 DIOPHANTUS OF ALEXANDRIA
V. 6. x-2 = r 2 ,
y-2
= s 2 , z-2
= t
2
,
yz — y — z = u 2 , zx — z — x = v 2 , xy
yz — x — w' 2 ,
zx — y — v' 2 , xy — z = w' 2 .
— x — y = %v 2 ,
[Solved by means of the proposition numbered (3) on
p. 481.]
Lemma 1 to V. 7. xy + x 2 -\-y 2 = u 2 .
(u 2 (v-
V. 7. x 2 ±{x + y + z)= , 2 , y 2 ±(x + y + z) =
,
(v' 2
z 2 ±{x + y + z)
= 'w2
w' 2
[Solved by means of the subsidiary problem (Lemma 2)
of finding three rational right-angled triangles with
equal area. If m, n satisfy the condition in Lemma 1,
i. e. if mn + on 2 + n 2 2
= p
, the triangles are formed ' ' from
the pairs of numbers (p, m), (p, n), (p,m + n) respectively.
Diophantus assumes this, but it is easy to prove.
In his case m = 3, n £= 5, so that p = 7. Now, in
a right-angled triangle,
(hypotenuse) 2 + four times area
is a square. We equate, therefore, x + y + z to four
times the common area multiplied by £
2
, and the several
numbers x, y, z to the three hypotenuses multiplied by £,
and equate the two values. In Diophantus's case the
triangles are (40, 42, 58), (24, 70, 74) and (15, 112, 113),
and 245£ = 3360£ 2 .]
V. 8. yz±(x + y + z)=
lu 2 (v 2
j
, 2 > zx±(x+y + z) = j
,2 >
or 4
2
+ 4m + 3 a + 1 = a square,
xy± (
x + y + z ) = |^2
.
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.]
[Solved by means of the same three rational rightangled
triangles found in the Lemma to V. 7, together
with the Lemma that we can solve the equations yz—a 2 ,
zx = b 2 2
, xy = c .]
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 ,
V. 9. (Cf. II. 11). x + y = 1, x + a = u 2 , y + a = v 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
V. 11. x + y + z= 1, x + a — u 2 , y + a=v 2 , z + a — w 2 .
[These are the problems of 7rapio-6Tr]Tos dyooyrj