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

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 ).}
([III. 15]. xy + x + ;>/
= u 2 , x+1 = —2 (y+l).
[III. 16]. xy — (x + y) = ti 2 , x—l
[IV. 32]. flj+l =^(aj_l).
=~z (y—\).
[V. 21]. x 2 + 1 = u 2 ,
y 2 + 1 = v 2 , s 2 + l = w 2 .
above.)
504 DIOPHANTUS OF ALEXANDRIA
IV. 8. Suppose % = £, 2/
3
= m 3 £ 3 ; therefore u=(m+ 1)£
must be the side of the cube m 3 £ 3 + £, and
m 3 £ 2 +l = (m 3 +3m 2 + 3m+l)f.
To solve this, we must have 3m 2 + 3m 4- 1
between consecutive cubes) a square.
(the difference
Put
3m 2 + 3m+l = (l—nm) 2 , and m = (3 -f 2n)/(n 2 — 3).
IV. 11. Assume x = (m+l)£, 2/
= ?^£> and we have
to make (3m 3 + 3m 2 + 1)£
2
equal to 1, i.e. we have
only to make 3m 2 + 3m + 1
IV. 18. x 3 + y = it
3
, 2/
2
+ # = v 2 .
IV. 24. a? + 2/
= a, fl?2/ = u 3 — u.
a square.]
[y = a — x; therefore ax — x 2 has to be made a cube
minus its side, say (mx— l) 3 — (mx— 1).
Therefore ax — x
2 = m3 & 3 — 3 m 2 a.* 2
+ 2 mx.
IV.
(IV.
(IV.
JV.
IV.
IV.
IV.
IV.
(viii)
Indeterminate analysis of the third degree.
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 .
8. x + y
3
= u 3 , x
+ y = u.
9. x + y
3 = 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 + y
2
= a cube, suppose z 2 + y
2
To make x 3 + y
2
x 3 .
a square, put # 3 = a 2 + b 2 ,
y 2 = 2 a&,
have then to make
which also satisfies x 3 2
— y = z 2 . We
2ab & square. Let a = g, b = 2g; therefore a 2 + b = 2 2
5 £
,
2
2a6 = 4| , 2/
= 2£, #:= £, and we have only to make
5£ 2 a cube. £ = 5, and ^ 3 = 125, / = 100, s 2 = 25.
To reduce this to a simple equation, we have only to
put m = |a.]
IV. 25. ^ + 2/-h0 = a, ^2/^ = { (
x —y) + (
ai — ^) + (2/ )
[The cube = 8(x — s)
3 .
3
*
(.a- > y > z)
Let x = (m+l)£,z = m£,so
that ?/ = 8£/(m 2 + m), and we have only to contrive that
8/(m 2 + m) lies between m and m + 1.
first limit 8 > m 3 + m 2 ,
8 = (m-f -|)
3
and puts
or m 3 + w 2 + |m + ^r ,
Dioph. takes the
whence m = §
; therefore & = §|, ?/ = §-£, £ = §£. Or >
multiplying by 15, we have x = 40 £, s/ = 27 £, = 25 £.
The first equation then gives £.]
rIV. 26. xy + x = u 3 , xy
llV. 27. xy — x = u z , xy
?J
IV. 28. xy + (x + y) = u , xy—(x
+ y = v ?> .
— y = v 3 .
+ y) = v\
[x + y = % (u 3 — v 3 ), ^2/ = i
('M' 3 + 3 ) j therefore
(x —
2
y)
= \ (u — 3 1> 3 2
)
— 2 (u3 + v 5 ),
which latter expression has to be made a square.