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I was reading Problem Solving Strategies by Arthur Engel (Chapter 6 p. 128) when I came across

$x^2 - dy^2 = 1\phantom{a}$ (1)

and

We associate the number $x+y\sqrt{d}$ with every solution $(x,y)$. We have the basic factorization $$x^2 - dy^2 = (x-y\sqrt{d})(x+y\sqrt{d}) \phantom{iii} (2)$$ It follows from (2) that the product or quotient of two solutions of (1) is again a solution of (1).

The way the book says this seems to suggest that this is a somewhat trivial observation, but I don't see it. I think I am aware of Brahmagupta's identity which says that the solutions are closed under multiplication but it seems that the algebra is somewhat involved. Can someone explain if there is a simple way to see how equation (2) shows that the solutions are closed under multiplication and division? Thanks.

hamburglar
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    FYI, from Wikipedia's History section of its "Pell's equation" article, there's "In Indian mathematics, Brahmagupta discovered that $(x_{1}^{2}-Ny_{1}^{2})(x_{2}^{2}-Ny_{2}^{2})=(x_{1}x_{2}+Ny_{1}y_{2})^{2}-N(x_{1}y_{2}+x_{2}y_{1})^{2}$". – John Omielan May 08 '24 at 21:20
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    simple way to show how solutions of (1) are "closed under multiplication": if $x_1^2-dy_1^2=1$ and $x_2^2-dy_2^2=1$ then $(x_1-y_1\sqrt d)(x_1+y_1\sqrt d)(x_2-y_2\sqrt d)(x_2+y_2\sqrt d)=1$, so $((x_1x_2+dy_1y_2)-(x_1y_2+y_1x_2)\sqrt d)((x_1x_2+dy_1y_2)+(x_1y_2+y_1x_2)\sqrt d)=1$, so $(x_1x_2+dy_1y_2)^2-d(x_1y_2+y_1x_2)^2=1$ – J. W. Tanner May 08 '24 at 21:33
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    I would not say "multiplication or division." The following things are true: there really is a "fundamental solution" $x_0^2 - n y_0^2 = 1$ where both $x_0, y_0 \neq 0$ and $|x_0| + | y_0 | $ minimal among nonzero solutions. Then $$ A = \left( \begin{array}{rr} x_0 & n y_0 \ y_0 & x_0 \ \end{array} \right) $$ or its inverse takes any solution to another; in the forward direction that is the mapping of pairs $(x,y) \mapsto (x_0 x + n y_0 y, y_0 x + x_0 y)$ – Will Jagy May 08 '24 at 21:34
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    Solutions are elements of $\Bbb Z[\sqrt d]$ of norm $1,$ so it follows immediately from multiplicativity of the norm: $N(\alpha)=1=N(\beta)\Rightarrow N(\alpha\beta) = N(\alpha)N(\beta)=1\cdot 1 = 1,$ where multiplicativity of the norm follows from multiplicativity of conjugation, i.e. $N(\alpha\beta)= \alpha\beta\overline{\alpha\beta} = \alpha\bar\alpha,\beta\bar\beta = N(\alpha)N(\beta).\ \ $ This question is - of course - a duplicate of many others. – Bill Dubuque May 08 '24 at 22:54

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