232 
THE ARITHMETICA 
Lemma I to the folloivingproblem. 
To find a right-angled triangle such that the difference of the 
perpendiculars is a square, the greater alone is a square, and further 
the area added to the lesser perpendicular gives a square. 
Let the triangle be formed from two numbers, the greater 
perpendicular being twice their product. 
Hence I must find two numbers such that (i) twice their 
product is a square and (2) twice their product exceeds 
the difference of their squares by a square. 
This is true of any two numbers the greater of which 
= twice the lesser. 
Form then the triangle from ;r, 2x, and two conditions are 
satisfied. 
The third gives 6,r 4 + t,x 2 = a square, or 6x 2 + 3 = a square. 
I have therefore to find a number such that 6 times its 
square + 3 = a square ; 
one such number is 1, and there are an infinite number of 
others * 1 . 
It x = 1, the triangle is formed from 1, 2. 
Suppose it formed from x+i, x; the sides then are 
f=2X 2 + 2X + X, £ = 2X+1, TJ — 2X 2 + 2X. 
i I 1 ^ J 
Thus - 4 . - & — x 4 + + 6x 2 + ^x + 1 - 4 [2X? + i,x 2 x) 
= x 4 - ^x 3 - 6x 2 + i 
= a square 
= {x 2 - 2x+ i ) 2 , say. 
Therefore - 6x 2 = 6x 2 - \x, x=~, and x+ i = - 
~~ « tlllU Ji -f 1 — ~ • 
3 3 
triangle (17, 15, 8). 
Take now 1 "¡x, 1 jx, 8a; for the sides of the triangle originally required to be found. 
We have then 
f+£--£ij= 32.*-6oa~ = 4 ; 
whence x—- 
3 
, and the required triangle is 
[The auxiliary right-angled triangle was of course necessary to be found in order to 
make the final quadratic give a rational result.] 
Bachet adds after vi. n a solution of the problem represented by 
2 
to which Fermat adds the enunciation of the corresponding problem 
£-±% v =a. 
2 
1 Though there are an infinite number of values of x for which 6x 2 + 3 becomes a square, 
the resulting triangles are all similar, h or, if x be any one of the values, the triangle is