A history of Greek mathematics - Wilbourhall.org

INDETERMINATE ANALYSIS 503
Let the first part = 5 z ; therefore f
= \\^ — z, or second part = 92 — Sz.
Therefore 5z + 92 - Sz = 72J, and z = \\\
(second part)
therefore the number of five-drachma measures is
\§ and
the number of eight-drachma measures ff .]
Lemma 2 to VI. 1 2. ax 2 + b = u 2 (where a + b — c
2). | / see p 457
Lemma to VI. 15. ax 2 -b=u 2 (where ad 2 -b = c 2 ).}
above.)
([III. 15]. xy + x + ;>/
=
u 2 , x+1
= —
2
(y+l).
[III. 16]. xy — (x + y) = ti 2 , x—l
=~z (y—\).
[IV. 32]. flj+l =^(aj_l).
[V. 21]. x 2 + 1 = u 2 ,
y 2 + 1 = v 2 , s 2 + l = w 2 .
(viii)
Indeterminate analysis of the third degree.
IV.
(IV.
(IV.
3. x 2 y — u, xy — u 3 .
6. x 3 + y
2
= u 3 , z 2 + y
2 = v
2
.
7. x 3 -\-y 2 = u 2 , z 2 + y
2 = v3 .
JV.
8. x + y
3
= u 3 , x
+ y = u.
IV.
IV.
IV.
IV.
3
9. x + y = u, x + y = u 3 .
10. x 3 -\-y = 3 x + y.
11. x 6 — y
6
— x — y.
12. 2^ + 2/
= y
3
+ x.,
the same problem.
(really reducible
to the second
degree.)
[We may give as examples the solutions of IV. 7
IV. 8, IV. 11.
IV. 7. Since z 2 2
+ y = a cube, suppose z 2 2
+ y
To make x 3 2
+ y
which also satisfies x 3 2
— y = z 2 . We
x 3 .
a square, put # 3 = a 2 + b 2 ,
y 2 = 2 a&,
have then to make
2ab & square. Let a = g, b = 2g; therefore a 2 + b 2 = 5 £
2
,
2a6 = 4|
2
, 2/
= 2£, #:= £, and we have only to make
5£ 2 a cube. £ = 5, and ^ 3 = 125, / = 100, s 2 = 25.