Problem source

\begin{questionparts}
\item Show that, if $\l a \, , b\r$  is \textbf{any} point on the curve $x^2 - 2y^2 = 1$, then $\l 3a + 4b \, , 2a + 3b \r\,$ also lies on the curve. 
\item Determine the smallest positive integers $M$ and $N$ such that, if $\l a \,,  b\r$ is \textbf{any} point on the curve $Mx^2 - Ny^2 = 1$, then $(5a+6b\,, 4a+5b)$ also lies on the curve.
\item Given that the point $\l a \, , b\r$ lies on 
the curve $x^2 - 3y^2 = 1\,$, find positive integers $P$, $Q$, $R$ and $S$ such that the point  $(P a +Q b\,, R a + Sb)$ also lies on the curve.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose $a^2-2b^2=1$ then

\begin{align*}
(3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\
&=a^2-2b^2 \\
&= 1
\end{align*}

Therefore $(3a+4b,2a+3b)$ also lies on the curve.

\item Suppose $Ma^2-Nb^2 = 1$ then

\begin{align*}
M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\
&= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2
\end{align*}
Therefore we need $3M = 2N$ so the smallest possible value would have to be $M = 2, N = 3$, which does work

\item Consider $x + \sqrt{3}y$, then consider $(x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}$. Notice that $(x+\sqrt{3}y)(x-\sqrt{3}y) = 1$ and $(2+\sqrt{3})(2-\sqrt{3}) = 1$ so $((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1$, so we can take $P=2,Q=3,R=1,S=2$
\end{questionparts}