A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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 + 2
+ Jam = 1 square]
)
m 2 + 1 = square)
m 2 + (2 a + 2) m + 1
solving in the usual way.]
VI. 10. \xy-\-x-\-z = a.
yi. 11. %xy-(x + z) = a.
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
and m is found.]
,
= { 1 + (a + 2) m — m 2 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.
510 DIOPHANTUS OF ALEXANDRIA
Lemma 1 .
If a rational right-angled triangle be formed
from m, n, the perpendicular sides are 2mn, m 2 — n 2 .
We will suppose the greater of the two to be 2mn.
The first two relations are satisfied by making m = 2 n.
Form, therefore, a triangle from £, 2£. The third condition
then gives 6 £
4
+ 3 £
2
= a square or 6 £2
+ 3 = a
square. One solution is £ = 1 (and there are an infinite
number of others to be found by means of it). If £ = 1,
the triangle is formed from 1, 2.
VI. 12. Suppose the triangle to be (kg, b£,p£). Then
, say, and £=p>/(k 2 —±pb).
(ipb)i 2 +pg = a, square = (k £)
2
This value must be such as to make (i2 j1j )£ 2 + b£ a square
also. By substitution of the value of £ we get
{bpk 2 + ±p 2 b(p-b)} /(P-ip6) 2 ;
so that bpk 2 + ^p 2 b(p—b) must be a square; or, if p,
the greater perpendicular, is made a square number,
bk 2 + ^pb(2J — b) has to be made a square. This by
Lemma 2 (see p. 467 above) can be made a square if
b + ipb{p — b) is a square. How to solve these problems,
says Diophantus, is shoivn in the Lemmas. It is not
clear how they were applied, but, in fact, his solution
is such as to make p, p — b, and b + ^j Jb all squares,
namely b = 3, p = 4, h = 5.
Accordingly, putting for the original triangle 3£, 4£, 5£,
we have
6 £
2
+ 4 £ = a square
6 £
2 -f 3 £ = a square
Assuming 6£ 2 + 4£ = m 2 £ 2 , we have £ = 4/(m 2 — 6), and
the second condition gives
or
96 12
m*-12m 2 + 36 + m^^6 ~ a SqUare '
1 2 m 2 + 2 4 = a square.
This can be solved, since m = 1 satisfies it (Lemma 2).
A solution is m 2 = 25, whence £ = T
4
§ .]
VI. 14. \xy — z — u 2 ,
\xy
— x — v 2 .
VI. 15. \xy-\-z = it 2 , -§#?/ + £' = v2 .