A history of Greek mathematics - Wilbourhall.org

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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.
Put
(the difference
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.
To reduce this to a simple equation, we have only to
put m = |a.]
IV. 25. ^ + 2/-h0 = a, ^2/^ = { (
x —y) + (
ai — ^) + (2/
)
3
*
(.a- > y > z)
[The cube = 8(x —
3 s) . 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 ,
and puts
Dioph. takes the
8 = (m-f -|)
3
or m 3 + w 2 + |m + ^r ,
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
+ y = v ?> .
llV. 27. xy — x = u z , xy
?J
IV. 28. xy + (x + y) = u , xy—(x
— 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.