Question

Let a and b be positive integers. Prove that if the equation $a{x}^2-b{y}^2=1$ has a solution in positive integers, then it has infinitely many solutions.

Answer 1

Factor the left-hand side to obtain $(x\sqrt{a}-y\sqrt{b})(x\sqrt{a}+y\sqrt{b})=1$ Cubing both sides yields $\left(\right(x^{3}a+3x{y}^{2}b)\sqrt{a}-(3x^{2}ya+y^{3}b\left)\sqrt{b}\right)\left(\right(x^3+3x{y}^2)\sqrt{a}+(3x^{2}ya+y^{3}b)\sqrt{b}=1.$ Multiplying out, we obtain $a(x^{3}a+3x{y}^{2}b{)}^2-b(3x^{2}ya+y^{3}b{)}^2=1$ Therefore, if $(x_1,y_1)$ is a solution of the equation, so is $(x_2,y_2)$ with $x_2=x_1^{3}a+3x_1{y}_1^2$ and $y_2=3x_1^2{y}_{1}a+y_1^{3}b.$ Clearly $,x_2>x_1$ and $y_2>y_1$.Continuing in this way we obtain infinitely many solutions in positive integers.