A history of Greek mathematics - Wilbourhall.org

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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
2 '
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.
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]