Show that, if \(\l a \, , b\r\) is any point on the curve \(x^2 - 2y^2 = 1\), then \(\l 3a + 4b \, , 2a + 3b \r\,\) also lies on the curve.
Determine the smallest positive integers \(M\) and \(N\) such that, if \(\l a \,, b\r\) is any point on the curve \(Mx^2 - Ny^2 = 1\), then \((5a+6b\,, 4a+5b)\) also lies on the curve.
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.
Solution:
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.
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
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 \(P=2,Q=3,R=1,S=2\)