Two problems on solving equations in rational and integers

Question

prove that $x^2-2y^2=7$ has infinitely many solutions in integers.

$a$ and $b$ are two non-zero rational numbers.Then if the equation $ax^2+by^2=0$ has a nonzero rational solution $(x_0,y_0)$ then for any rational $t$, $ax^2+by^2=t$ also has a non zero rational solution $(x_t,y_t)$.Prove this.

Answer 1

Define the sequences $(x_n,y_n)$ recursively by ($x_1=3$ and $y_1=1$) \begin{eqnarray*} x_{n+1}=3x_n+4y_n \\ y_{n+1}=2x_n +3y_n. \end{eqnarray*} It easy to show by induction that the pairs $(x_n,y_n)$ will satisfy $x^2-2y^2=7$. \begin{eqnarray*} x_{n+1}^2-2y_{n+1}^2=(3x_n+4y_n)^2-2( 2x_n +3y_n)^2 \\ =9x_{n}^2+24x_ny_n+16y_{n}^2-2(4x_{n}^2+12x_ny_n+9y_{n}^2)=x_{n}^2-2y_{n}^2=7. \end{eqnarray*}

Answer 2

The first equation is what is generally known as the Pell's equation. For this it is easy to see that $(3,1)$ is a solution. Then one can easily check that any $(x_n, y_n)$ satisfying $$x_n+\sqrt{2}y_n=(3+\sqrt{2})(3+2\sqrt{2})^n$$ will also be solutions to the given equation, hence infinitely many integer solutions.

For the second question, please share your thoughts so that it will be easier to see where you need help.