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

INDETERMINATE ANALYSIS 495
III. 14. yz + x 2 = u 2 , 2x + y
2 = v
2, xy + z 2 — w 2 .
III. 15. yz+(y + z) = u 2 , zx + (z + x) = v 2 , xy
+ (x + y) = w 2 .
[Lemma. If a, a+l be two consecutive numbers,
a 2 (a + l) 2 + a 2 + (a + I) 2 is a square. Let
7/ = m 2 , z = (m + 1 )
2
.
Therefore (m 2 + 2m + 2) a; + (m + 1
and (m 2 +l)o? + m 2
have to be made squares. This is solved as a doubleequation
;
in Diophantus's problem m = 2.
Second solution. Let x be the first number, m the
second; then (m+l)x + m is a square = n 2 , say; therefore
x = (n 2 — m)/(m+ 1), while y = m. We have then
(m + 1) + m = a square
/n 2 + l\ n 2 — m
and (
—- ) z + — — = a square
\m + 1 / -m-H 1
Diophantus has m = 3, n = 5, so that the expressions
to be made squares are with him
This is
42 + 3 \
6iz + 5i\
not possible because, of the corresponding coefficients,
neither pair are in the ratio of squares.
2 '
In order to
substitute, for 6 J, 4, coefficients which are in the ratio
of a square to a square he then finds two numbers, say,
p }
q to replace 5^, 3 such that pq+p + q = a square, and
(p + 1 ) / (q + 1 ) = a square. He assumes £ and 4 £ + 3
which satisfies the second condition, and then solves for g }
which must satisfy
4 £
2
+ 8 £ + 3 = a square = (2£ -3) 2 ,
say,
which gives £ = t
3 q, 4£ + 3 — 4-|.
He then solves, for z, the third number, the doubleequation
5~z + 4-§- = square]
To^ + T
3o = square]
496 DIOPHANTUS OF ALEXANDRIA
after multiplying by 25 and 100 respectively, making
expressions
130^+105)
130.T+
30J
In the above equations we should only have to make
n 2 + 1 a square, and then multiply the first by n 2 + 1 and
the second by (m + l) 2 .
Diophantus, with his notation, was hardly in a position
to solve, as we should, by writing
(y + i)(z + i)=a 2 +l,
(Z + l)(X +l) = b 2 +l,
(x+l)(y+l) = c 2 +l,
which gives x + 1 = V { (b 2 + 1) (c 2 + I) /(a 2 + 1) }, &c]
III. 16. yz — (y + z) = u 2 ,
zx—(z + x) = v 2 , xy—{x + y) = w 2 .
[The method is the same mutatis mutandis as the
second of the above solutions.]
rill. 1 7. xy + (x + y) = u 2 , xy + x = v 2 ,
xy + y = w 2 .
llll. 18. xy — (x + y)= u 2 , xy — x = v 2 ,
xy — y — iv 2 .
III. 1 9. (x x
+ x 2
+ x. 6
+ x^) 2 ± x x
=
^&'i -f- Xa ~r «£•> ~r "-'4/ jt *^s
"~
1
^t^j T ^'2 ' ^3
j
(t 2
I
It/
IT
'2
" ^4) i — "" *^4 z