A history of Greek mathematics - Wilbourhall.org

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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
the second condition gives
have £ = 4/(m 2 — 6), and
96 12
m*-12m 2 + 36 + m^^6 ~ a SqUare '
or
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 .