A history of Greek mathematics - Wilbourhall.org

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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
=
j
(t 2
I It/
IT
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