The Pell equation, by Edward Everett Whitford.

THE PELL EQUATION

37

negative right members according to convenience. The
following is Bhaskara's solution of the problem just mentioned, taken from Strachey's1 translation of a Persian
manuscript of 1634, which is considerably clearer than
the wording of Colebrooke.
"What square is that which being multiplied by 61,
and the product increased by 1, will be a square? Let 1
be the less root; 8 is the greater; and 3 the augment
affirmative. Applying the operation of the multiplicand
it is thus:
Dividend Divisor Augment
1 3 8
Reject the divisor twice from the augment, 2 remains;
and after the operation 2 the multiplicand, and cipher the
quotient are obtained. As the line is odd we subtract
cipher from the dividend and 2 from the divisor. It is 1
and 1. As we reject the divisor twice from the augment,
we add 2 to the quotient. The quotient is 3 and the multiplicand 1. If we subtract the square of the multiplicand,
which is 1, from 61, a greater number remains. We therefore add twice the dividend and the divisor to the quotient
and the multiplicand. The quotient is 5 and the multiplicand 7."
All these words are used to explain how the value of rl
in (3) is obtained. The solution continues:
"Subtract the square of 7 from 61; 12 remains. Divide
by the augment of the operation of multiplication of the
square which is 3 affirmative; 4 affirmative is the quotient;
and after revision it is 4 negative; and this the augment;
and the quotient which was 5 is the less root; 39 then will
be the greater root. As 4 is not the original augment we
have found 2 an assumed number, and by its square we
divided this augment. 1 the augment negative is the
quotient. We also divide 5 and 39 by 2. These same
two numbers, with the denominator 2, are the quotients.
E. Strachey, p. 46.