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Usually the Pell equation is written $x^2 - dy^2 = 1$ but here I am looking for solutions to an equation of the type: $$ x^2 - k xy + y^2 = 1 $$ and In particular, $k$ is a perfect square. So I am picking $k = 25$ an example.

If we complete the square then $25/2$ is not an integer.
$$ xy + x^2 - 26xy + y^2 = x^2 + xy + \big(x - 13y\big)^2 = 170$$ I think I am better of solving the orginal problem. Can any variant on the Pell equation work?

cactus314
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  • yes, solve it as quadratic equation in x or y, this should lead to a Pell equation. For example (624,15575) is a solution. – i9Fn Nov 15 '16 at 15:19
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    $4x^2-4kxy + 4y^2=4$. Complete the square, $(2x-ky)^2-(k^2-4)y^2=4$ is your pell-like equation. You need $u^2-(k^2-4)v^2=4$ and you need $u+kv$ even. – Thomas Andrews Nov 15 '16 at 15:24
  • In particular, $(x,y)=(k,1)$ and $(x,y)=(1,k)$ are solutions to your original equation, (as are $(1,0)$ and $(0,1)$.) – Thomas Andrews Nov 15 '16 at 15:31

2 Answers2

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What you do is "depress" the equation (get rid of the $x^{n-1}$ term), the same trick used by Vieta to solve the general cubic. Given, $$x^2-k xy+y^2=1$$ Let $x=u+av,\,$ and $y=bv$ to get, $$u^2 + (2 a - b k) u v + (a^2 + b^2 - a b k) v^2=1$$ Then just choose integers $a,b$ such that $2 a - b k=0$, and if $a^2 + b^2 - a b k<0$, then you'll get a Pell equation in standard form. For yours, with $a=25,b=2,$ what you get is, $$u^2-621v^2=1$$ hence has infinitely many solutions.

Tito Piezas III
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Your discriminant, for $x^2 - k x y + y^2,$ is $$ \Delta = k^2 - 4. $$ The smallest solution, in positive integers, to $$ \tau^2 - \Delta \sigma^2 = 4 $$ is $\tau = k, \sigma = 1.$ You have the obvious $-1$ automorphism given by interchanging $x,y.$ The matrix generating the oriented automorphisms is $$ \left( \begin{array}{rr} \frac{\tau - B \sigma}{2} & - C \sigma \\ A \sigma & \frac{\tau + B \sigma}{2} \end{array} \right). $$ With $A = 1, B = -k, C = 1$ this becomes $$ P = \left( \begin{array}{rr} k & - 1 \\ 1 & 0 \end{array} \right). $$ $$ P^{-1} = \left( \begin{array}{rr} 0 & 1 \\ -1 & k \end{array} \right). $$ Beginning with a column vector with entries $(1,0),$ multiplying $P$ times the column repeatedly gives (columns) $(1,0),$ $(k,1),$ $(k^2 - 1,k),$ $(k^3 - 2k,k^2 - 1),$ and so on forever.

as in $$ \left( \begin{array}{r} x_{n+1} \\ y_{n+1} \end{array} \right) = \left( \begin{array}{rr} k & - 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{r} x_n \\ y_n \end{array} \right) $$

From Cayley-Hamilton on the matrix $P$, we get that both $x_n$ and $y_n$ sequences obey linear recursions $$ x_{n+2} = k x_{n+1} - x_n, $$ $$ y_{n+2} = k y_{n+1} - y_n. $$ For this particular problem, this is also evident from the matrix $P$ since the $x$ and $y$ sequences are the same, only $y_{n+1}= x_n.$

Will Jagy
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