Looking at an old exam paper I cam across a question which goes as follows :
1) Show that the continued fraction of $\sqrt{14}=[3;\overline{1,2,16}]$.
2)hence or otherwise obtain 3 solutions to $x^2-14y^2=1$
3) Determine all $a,b \in \Bbb Z$ satisfying the Diophantine equation $ax^2_ 0 + by^2_ 0 = 1$ for $(x_0,y_0)$ as in (2)
4) Let $(xi,yi)$ be the $i^{th}$ solution to the Diophantine equation in part (2). Explain why solutions $a,b \in \Bbb Z$ always exist to the Diophantine equation $ax^2 _i +by^2_i = 1.$
1) and 2) are trivial but 3 and 4 have me quite stumped I looked back at my old notes and I couldn't find anything about it. Could anyone provide method of proof for 3) and an explanation for 4) ?
Note: I only want to use elementary techniques of number theory as I'm only just learning it, and doing so in the context of number theory .
HINT.
►The fondamental unit of the field $\mathbb Q(\sqrt{14})$ is easily find (besides of the continued fraction) with the first $b^2$ such that $14b^2$ be an square minus $1$ so we determine from $14\times(4^2)=15^2-1$ that all the solutions of $x^2-14y^2=1$ are given by $$a_n+b_n\sqrt{14}=(15+4\sqrt{14})^n$$ ►If $x_i^2-14y_i^2=1$ and we want to solve $ax_i^2+by_i^2=1$ (where the unknowns are $a$ and $b$) it is clear that $(a,b)=(1,-14)$ is a solution. And it is quite known the parameterization of all the solutions (infinitely many) of a linear equation when we have a particular solution.