250 dhruv raina
In modern language this could be translated as a- 3 − 1 = c − b , where
Bhaskara immediately suggests taking c − b as 2. In this demonstration
the diff erence between one of the sides (upright) and the hypotenuse is
assumed as 2.
(a) Th e square of that added to one to which 3 is added: (2 + 1)^2 + 3 = 12 – this is the
side.
(b) 12^2 = 144 – this is the diff erence between the squares of the hypotenuse and side
(upright).
By the rule the diff erence of the squares is equal to the product of the sum and
diff erence
Which means a 2 − b 2 = ( a + b )( a − b ).
It is in this context that here Bhaskara includes a proof of the rule, to
which Colebrooke refers. Th is proof as is evident is based on a form of rea-
soning that draws upon fi gures with particular dimensions. Th e text then
gives the square of 7 as 49 represented as below ( Figure 5.1 ):Figure 5.1 Th e square a^2.From this square of 7 × 7 subtract a square of 5, which is 25.
Th is gives the following ( Figure 5.2 ).
We are left with a remainder of 24.
a − b = 2 and a + b =12 and the product consists of 24 equal cells
( Figure 5.3 ).