Page:A History of Mathematics (1893).djvu/115

 Euler. Euder's process of reducing $$\scriptstyle{\frac{a}{b}}$$ to a continued fraction amounts to the same as the Hindoo process of finding the greatest common divisor of a and b by division. This is frequently called the Diophantine method. Hankel protests against this name, on the ground that Diophantus not only never knew the method, but did not even aim at solutions purely integral.[7] These equations probably grew out of problems in astronomy. They were applied, for instance, to determine the time when a certain constellation of the planets would occur in the heavens.

Passing by the subject of linear equations with more than two unknown quantities, we come to indeterminate quadratic equations. In the solution of $$\scriptstyle{xy=ax+by+c}$$, they applied the method re-invented later by Euler, of decomposing $$\scriptstyle{(ab+c)}$$ into the product of two integers $$\scriptstyle{m \cdot n}$$ and of placing $$\scriptstyle{x=m+b}$$ and $$\scriptstyle{y=n+a}$$.

Remarkable is the Hindoo solution of the quadratic equation $$\scriptstyle{cy^2=ax^2+b}$$. With great keenness of intellect they recognised in the special case $$\scriptstyle{y^2=ax^2+1}$$ a fundamental problem in indeterminate quadratics. They solved it by the cyclic method. "It consists," says De Morgan, "in a rule for finding an indefinite number of solutions of $$\scriptstyle{y^2=ax^2+1}$$ (a being an integer which is not a square), by means of one solution given or found, and of feeling for one solution by making a solution of $$\scriptstyle{y^2=ax^2+b}$$ give a solution of $$\scriptstyle{y^2=ax^2+b^2}$$. It amounts to the following theorem: If p and q be one set of values of x and y in $$\scriptstyle{y^2=ax^2+b}$$ and $$\scriptstyle{p'}$$ and $$\scriptstyle{q'}$$ the same or another set, then $$\scriptstyle{qp+pq'}$$ and $$\scriptstyle{app'+qq'}$$ are values of x and y in $$\scriptstyle{y^2=ax^2+1}$$. From this it is obvious that one solution of $$\scriptstyle{y^2=ax^2+1}$$ may be made to give any number, and that if, taking b at pleasure, $$\scriptstyle{y^2=ax^2+b^2}$$ can be solved so that x and y are divisible by b, then one preliminary solution of $$\scriptstyle{y^2=ax^2+1}$$