A history of Greek mathematics - Wilbourhall.org

498 DIOPHANTUS OF ALEXANDRIA
Similarly we satisfy the second condition by assuming
zx — n
2
£
2
+ 2ng ; therefore x = n 2 g + 2 n. To satisfy the
third condition, we must have
(m 2 n 2 £ 2 + 2mn .
m
+ n£ + 4m%) + 1 a square.
We must therefore have 4 mn + 1 a square and also
mn(m + n) = mn V(4mn-\- 1). The first condition is
satisfied by n = m-\- 1 , which incidentally satisfies the
second condition also. We put therefore yz = (m£ + l) 2 — 1
2
and zx— {
(m + l)£ + 1 }
— 1, and assume that z = g, so that
y = m 2 g + 2m, x = (m + 1) 2 £ + 2(m + 1), and we have
shown that the third condition is also satisfied.
Thus we
have a solution in terms of the undetermined unknown £.
The above is only slightly generalized from Diophantus.]
IV. 20. x 2
x z
-\- 1 = r 2 , x
X-i X/i ~f~ 1 — iXi
z
x Y
-\~ 1 = s 2 ,
x
, Xc) Xi ~t~ J- —
1
x 2
+ 1 = t 2 ,
U ,
Xr, Xa ~t" i — w .
[This proposition depends on the last, x lt
x 2
, x z
being
determined as in that proposition. If x z
corresponds to z
in that proposition, we satisfy the condition x 3
x 4
+l = w 2
by putting x 3
x± = {(m + 2)£ + 1 }
2 — 1, and so find x 4
in
terms of £, after which we have only two conditions more
to satisfy. The condition x 1
x + 1 = square is automatically
satisfied,
4:
since
f
(m + 1)
2
£ + 2 (m + 1)} {
(m + 2)
2
£ + 2 (m + 2) } + 1
is a square, and it only remains to satisfy x 2
x i
+l = square.
That is,
(m 2 |+2m) {(m + 2)
2
£ + 2(m + 2)} + 1
= m 2 (m+2) 2 f + 2m(m + 2)(2m + 2)^ + 4m(m+2) + 1
has to be made a square, which is easy, since the coefficient
of £
2
is a square.
With Diophantus m = 1, so that x x
= 4£ + 4, x 2
= £ + 2,
# 3
= £, # 4
= 9£ + 6, and 9£ 2 + 24£+13 has to be made
a square. He equates this to (3£— 4) 2 ,
giving £ = y
1^.]
IV. 21. xz = y
2
, x — y = u 2 , x — z = v 2 , y
— z — w 2 . (x>y>z)
IV. 22. xyz + x = u 2 , xyz + y — v 2 , xyz + z = w 2 .
IV. 23. xyz — x = u 2 , xyz — y = i;
2
,
xyz — z = iv 2 .