INDETERMINATE ANALYSIS 
509 
(VI. 8. %xy + (x + y) — a. 
(VI. 9. ^xy—{x + y) = a. 
[With the same assumptions we have in these cases 
to make {^{p + h)} 2 + a{^pb) a square. Diophantus 
assumes as before 1, m for the values of p, h, and obtains 
the double equation 
\ (m + 1 ) 2 + |am = square [ 
m 2 + 1 = square) 
m 2 + (2 a + 2) m + 1 = square) 
or 1 5 
m 2 + 1 = square) 
solving in the usual way.] 
VI. 10. \xy + x-rz = a. 
,VI. 11. \ xy~{x + z) = a. 
[In these cases the auxiliary right-angled triangle has 
to be found such that 
{\ (J l +^)} 2 + «■ (ipb) = a square. 
Diophantus assumes it formed from 1, m + 1; thus 
%{h+pf = £{m 2 + 2m + 2 +m 2 + 2 m} 2 = (m 2 + 2m+ l) 2 , 
and a{%pb) = a{m+ 1) (m 2 + 2m). 
, Therefore 
m 4 + (<x + 4)m 3 + (3a + 6)m 2 + (2a + 4)m+ 1 
= a square 
= {1 + (a + 2) m — m z } 2 , say; 
and m is found.] 
Lemma 1 to VI. 12. x = u 2 , x—y = v*, y + y = to 2 . 
[VI. 12. \xy + x — U 2 , \xy + y = v 2 . 
IVI. 13. \xy — x = u 2 , \xy — y = v 2 . 
[These problems and the two following are interesting, 
but their solutions run to some length; therefore only 
one case can here be given. We will take VI, 12 with 
■ its Lemma 1.