BOOK VI 
23T 
Given number 4. 
If we assumed as the triangle (bx, px, bx), we should have 
\pbx 2 + hx + bx = 4; 
and, in order that the solution may be rational, we must 
find a right-angled triangle such that 
l (hyp. + one perp.) 2 -f 4 times area = a square. 
Form a right-angled triangle from 1, m + 1. 
Then \ (hyp. + one perp.) 2 = a (m 2 4- 2m + 2 + m 2 + 2mf 
= m 4, + 4m 3 -l- 6 m 2 + 4 m + 1, 
and 4 times area = 4 (m + i)(m 2 + 2m) 
— 4m 3 + 12 m 2 + 8 m. 
Therefore 
m 4 + 8m 3 + 1 Sm 2 + 12m + 1 = a square = (6m +1 — m 2 ) 2 , say, 
whence w = f, and the auxiliary triangle is formed from 
(1, |) or (5, 9). This triangle is (56, 90, 106) or 
(28, 45, 53). 
We assume therefore 2Sx, 45^, 53.tr for the original triangle, 
and we have 630.tr 2 + 8i.r = 4. 
Therefore x = T ^, and the problem is solved. 
11. To find a right-angled triangle such that its area minus 
the sum of the hypotenuse and one of the perpendiculars is a given 
number. 
Given number 4. 
We have then to find an auxiliary triangle with the same 
property as in the last problem ; 
therefore (28, 45, 53) will serve the purpose. 
We put for the triangle of the problem (28^,45^, 53-^), and 
we have 630.tr 2 — Six — 4 ; 
x = and the problem is solved 1 . 
1 Diophantus has in vi. 10, n shown us how to find a rational right-angled triangle 
f, £, 7j (f being the hypotenuse) such that 
(0 \& + =a > 
(2) +£) = «. 
Fermat, in the Inve.ntum Novum, Part in. paragraph 33 (Oeuvres de Fermat, in. 
p. 389), propounds and solves the corresponding problem 
(3) = 
In the particular case taken by Fermat a = 4. He proceeds thus: 
First find a rational right-angled triangle in which (since a = 4) 
I2 ^ + *> 1 “ 4 • \ = a square.