A history of Greek mathematics - Wilbourhall.org

VI. 8. \xy +(x + y) = a.
VI. 9.
INDETERMINATE ANALYSIS 509
ixy-(x + y) =a.
[With the same assumptions we have in these cases
to make {%{
rp + b)} 2 + a{^pb) a square. Diophantus
assumes as before 1 m , for the values of p, b, and obtains
the double equation
or
J (m + 1 )
m 2 + (2 a + 2) m + 1
solving in the usual way.]
VI. 10. \xy-\-x-\-z = a.
yi. 11. %xy-(x + z) = a.
2
+ Jam = square]
m 2 + 1 = square)
m 2 + 1
= square]
= square)
[In these cases the auxiliary right-angled triangle has
to be found such that
{ i Q 1 + V)
}
2
+ a (i pty — a square.
.
;
Diophantus assumes it formed from 1 , m
+ 1 ;
thus
| (A +p) 2 = J [m 2 + 2m + 2 + m 2 + 2m} 2 = (m 2 + 2m + l) 2 ,
and a (^ £>6) = a (m + 1 )
(m 2 + 2 ??i)
Therefore
m 4 + (a + 4)m 3 + (3a + 6)m 2 + (2a + 4)m + 1
= a square
= { 1 + (a + 2) m — m 2 }
and m is found.]
,
2
, say
Lemma 1 to VI. 12. x = u 2 , x — y — v 2 \xy
,
+ y = w 2 .
fVI. 12. i 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. 1 2 with
its Lemma 1.