RATIONAL RIGHT-ANGLED TRIANGLES 
81 
Another formula, devised for the same purpose, is attributed 
to Plato, 1 namely 
(2 m) 2 + (m 2 — l) 2 = (m 2 + l) 2 . 
We could obtain this formula from that of Pythagoras by 
doubling the sides of each square in the latter; but it would 
be incomplete if so obtained, for in Pythagoras’s formula m is 
necessarily odd, whereas in Plato’s it need not be. As Pytha 
goras’s formula was most probably obtained from the gnomons 
of dots, it is tempting to suppose that Plato’s was similarly 
evolved. Consider the square with n dots in its 
side in relation to the next smaller square (n — 1 f ■ *— 
and the next larger {n +1} 2 . Then n 2 exceeds ..... 
(n— l) 2 by the gnomon 2u,—1, but falls short of » • • • • 
(n+ l) 2 by the gnomori 2n + 1. Therefore the * 
square (n+1) 2 exceeds the square (n~ l) 2 by 
the sum of the two gnomons 2/1—1 and 2n + l, which 
is 4 n. 
That is, 4n + (n— l) 2 = (n+ l) 2 , 
and, substituting m 2 for n in order to make 4 a square, we 
obtain the Platonic formula 
(2m) 2 + (m 2 — l) 2 — (m 2 + l) 2 . 
The formulae of Pythagoras and Plato supplement each 
other. Euclid’s solution (X, Lemma following Prop. 28) is 
more general, amounting to the following. 
If iil be a straight line bisected at G and produced to D, 
then (Eucl. II. 6) 
AD.DB + GB 2 ^ CD 2 , 
which we may write thus: 
uv — c 2 — b 2 , 
where u = c + b, v = c — b, 
and consequently 
c=-\{u + v), 6=i(u —v). 
In order that uv may be a square, says Euclid, u and v 
must, if they are not aqjually squares, be ‘ similar plane num 
bers ’, and further they must be either both odd or both even 
Proclus on Eucl. I, pp.428. 21-429. 8.