A history of Greek mathematics - Wilbourhall.org

a
—
m
INDETERMINATE ANALYSIS 513
a cube. Put 2£ 2 + 2£ = m 2 | 2 , so that £ = 2/(m 2 -2),
and we have to make
8 8 2 2 m, 4
-, +
(m 2 -2) 3 (m /
2 -2) m 2 2 -2 (m 2 —2) 3
Make 2m .
o—t^to + —o— ~' or ;— 5—^ ,
cube = n^, so that 2m 4 = m 3^ 3,
a cube.
and
g
m = ^n 3 : therefore £= —
> and £ must be made
greater than 1 , in order that £
2
— 1 may be positive.
Therefore 8 < n G < 16;
this is satisfied by n- = G - 7 9 5
¥\
- or w =
- 2
g 7 -,
and m = f|.]
VI. 22. x + y + z = u 3 ,
%xy + (x + y + z) = v 1 .
[(1) First seek a rational right-angled triangle such
that its perimeter and its area are given numbers,
say p, m.
Let the perpendiculars be -, 2 m £; therefore the hypoi
tenuse = p — - — 2m£, and (Eucl. I. 47)
2 + 4m2 £ 2 + (p
2
+ 4 m)
f— 4mpg — — + 4m
2
£
2
,
2 /j
or p 2 + 4 m = 4 mp£ + -~- >
that is,
(jo 2 + 4m)£ = ^mpg* + 2^>.
(2) In order that this may have a rational solution,
2 2
{ i {'P + 4m) — } %p m l must be a square,
i.e. 4
2
— 6^)
2m -f ^_p
4
= a square,
or m 2 — §p
2m + t
1©^4 = a square]
Also, by the second condition, m+p = a square)
To solve this, we must take for p some number which
is both a square and a cube (in order that it may be
possible, by multiplying the second equation by some
square, to make the constant term equal to the constant
1523.2 \j 1