INDETERMINATE ANALYSIS 
511 
[The auxiliary right-angled triangle in this case must 
be such that 
m 2 hp — \pb. p (h —p) is a square. 
If, says Diophantus (VI. 14), we form a triangle from 
the numbers X x , X 2 and suppose that p = 2X 1 X 2 , and if 
we then divide out by (X 1 — X 2 ) 2 , which is equal to h~p, 
we must find a square k 2 [ = m 2 / (X 1 — X 2 ) 2 ] such that 
k 2 Up — \pb .p is a square. 
The problem, says Diophantus, can be solved if X x , X 2 
are ‘ similar plane numbers 5 (numbers such as ab, — ab). 
This is stated without proof, but it can easily be verified 
that, if k 2 = X x X 2 , the expression is a square. Dioph. 
takes 4, 1 as the numbers, so that k 2 = 4. The equation 
for m becomes 
8. 17m 2 —4. 15.8.9 = a square, 
or 136m 2 — 4320 = a square. 
The solution m 2 = 36 (derived from the fact that 
k 2 = m 2 /{X x — X 2 ) 2 . or 4 — m 2 /3 2 ) 
satisfies the condition that 
m 2 hp—%pb . p {h —p) is a square.] 
VI. 16. £ + r] = x, £/rj = y/z. 
[To find a rational right-angled triangle such that the 
number representing the (portion intercepted within 
the triangle of the) bisector of an acute angle is rational. 
Let the bisector be 5£, the segment BD of the base 3£, 
so that the perpendicular is 4£. 
Let CB = 3 n. Then AC : AB = CD : DB,