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 m, 4 
2 
(m 2 — 2) 3 "** (m 2 —2) 2 + m 2 — 2’ 01 
Make 2 m a cube — n 3 , so that 2 m 4 = m 3 n 3 , and 
g 
m = hrP; therefore £ = ———, and £ must be made 
* b n b —8 s 
greater than 1, in order that ¿ 2 — 1 may be positive. 
Therefore 
8 < n G < 16; 
this is satisfied by n 6 = or n 3 = - 2 g 7 -, and m = .] 
VI, 22. x+y+z— u 3 , \xy + (x + y + z) = v 2 . 
[(1) First seek a rational right-angled triangle such 
that its perimeter and its area are given numbers, 
say p, m. 
Let the nernendiculars be -. 2 m P: therefore the hvpo- 
or 
that is, 
(p 2 + 4 m) £ = 4 mp + 2 p. 
(2) In order that this may have a rational solution, 
{\ ip 2 + 4 m) } 2 — 8p 2 m must be a square, 
4m 2 —Gp 2 m + %p i = a square, 
i.e. 
or 
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 
1$23.2 Xi 1