INDETERMINATE EQUATIONS 
471 
Ex, from Diophantus : 
65— 6« = u 2 
65 — 24« = w 2 
(IV. 32) 
therefore 
260-24« = u' 2 1 
65 — 24« =iv' 2 ) 
The difference = 195 = 15.13, say; 
therefore ^(15 — 13) 2 = 65 — 24«; that is, 24« = 64, and« = f. 
Taking now tlic condition (2) that ah is a square, we see 
that the equations can be solved in the cases where either 
a and h are both squares, or the ratio of a to h is the ratio of 
a square to a square. If the equations are 
0C«+ c 2 = u 2 , 
/3« + d 2 = w 2 , 
and factors are taken of the difference between the expressions 
as they stand, then, since one factor p, as we saw, satisfies the 
{p 2 — (c 2 + d 2 ) } 2 = 4 c 2 d 2 . 
p = c±d. 
equation {p 2 
we must have 
Ex. from Diophantus : 
(III. 15) 
The difference is 5«+5 = 5(«+l); the solution is given by 
(f« + 3) 2 = 10« + 9, and « = 28. 
Another method is to multiply the equations by squares 
such that, when the expressions are subtracted, the absolute 
term vanishes. The case can be worked out generally, thus. 
Multiply by d 2 and c 2 respectively, and we have to solve 
ocd 2 x + c 2 d 2 = u 2 | 
/3 c 2 « + c 2 d 2 = id 2 1 
Difference = {ad 2 — /3c 2 )« = px ,q say. 
Then « is found from the equation 
ad 2 x + c 2 d 2 = \ {px + q) 2 , 
which gives p 2 x 2 + 2«{pq — 2ad 2 ) + q 2 —4c 2 d 2 = 0,