HERONIAN INDETERMINATE EQUATIONS 445 
number 3 is of course only an illustration, and the problem is 
equivalent to the solution of the equations 
(1) u + v = n(x + y) | 
(2) xy — n.uv j 
The solution given in the text is equivalent to 
x = 2ft 3 - 1, y — 2 n 3 ) 
u = n{4:n z —2), v = n j 
Zeuthen suggests that the solution may have been obtained 
thus. As the problem is indeterminate, it would be natural 
to start with some hypothesis, e.g. to put v — n. It would 
follow from equation (1) that u is a multiple of n, say nz. 
We have then 
x + y = 1 + z, 
while, by (2), xy = n 3 z, 
whence xy = n 3 {x + y) — n 3 , 
or (x — n 3 ) {y — n 3 ) — n 3 (n 3 — 1 ). 
An obvious solution is 
x — n 3 = n 3 — 1, y — n 3 = n 3 , 
which gives z — 2n 3 —l + 2n 3 — 1 = 4% 3 — 2, so that 
u = nz — n(4n 3 —2). 
II. The second is a similar problem about two rectangles, 
equivalent to the solution of the equations 
(1) x + y = u + v) 
(2) xy =n.uvj 
and the solution given in the text is 
x + y = u + v = n 3 —l, (3)