A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$

Question

A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$

313 Views Asked by Bumbble Comm At 25 Mar 2026 - 8:36

Find all the pairs of positive integers $(x,y)$ satisfying $$ x^2 + y^2 - 5xy + 5 = 0 . $$

I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.

For example, the first general solution presented reads:

$$ \begin{array} { r c l } x &=& \dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n - \sqrt{21} \left(55 - 12\sqrt{21} \right)^n + 21 \left(55 + 12\sqrt{21} \right)^n + \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\ y &=& \dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n - 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n + 63 \left(55 + 12\sqrt{21} \right)^n + 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.

Plugging $n=0,1,2,3$ gives the first few solutions

$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449) $$

I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.

Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?

Original Q&A

There are 4 best solutions below

Bumbble Comm On 19 Jan 2021 - 5:08

It's $(5y-2x)^2-21y^2=-20$, which is a Pell type equation.

I got that by completing the square:

$x^2-5xy+y^2=-5\implies \left(x-\frac52y\right)^2-\frac{21}4y^2=-5\implies (5y-2x)^2-21y^2=-20$.

So it's $X^2-21Y^2=-20$, with $X=5y-2x$ and $Y=y$.

Bumbble Comm On 19 Jan 2021 - 7:42

The equation is symmetric and it is easy to see solutions if, for example, we solve for $y$.

$$x^2 + y^2 - 5xy + 5 = 0 \implies\quad y = \frac{5 x \pm \sqrt{21 x^2 - 20}}{2}\qquad |x|\ge 1$$

Note that the absolute value of $x$ must be at least $1$ for the radical to be non-negative and therfore, for $y$ to be real.

Given this equation, we can also see that there are $2$ $y$-values for every valid $x$. There are $28$ solutions for $\space -50000\le x \le 50000.\quad$ Here is that "sample" of $\quad (x,y_1,y_2.)\quad$

$$ (-7369,-35307,-1538)\quad (-4729,-22658,-987)\quad (-1538,-7369,-321)\\ (-987,-4729,-206)\qquad (-321,-1538,-67)\qquad (-206,-987,-43)\\ (-67,-321,-14)\qquad (-43,-206,-9)\qquad (-14,-67,-3)\\ (-9,-43,-2)\quad (-3,-14,-1)\quad (-2,-9,-1)\quad (-1,-3,-2)\\ (1,2,3)\qquad (2,1,9)\qquad (3,1,14)\qquad (9,2,43)\qquad (14,3,67)\\ (43,9,206)\qquad (67,14,321)\qquad (206,43,987)\qquad (321,67,1538)\\ (987,206,4729)\qquad (1538,321,7369)\qquad (4729,987,22658)\\ (7369,1538,35307)\quad (22658,4729,108561)\quad (35307,7369,169166)$$

Note that all the negative $x$-values have positive counterparts and that we have a counterpart solution for $x$.

$$x = \frac{5 y \pm \sqrt{21 y^2 - 20}}{2}$$ so the $x,y$ values should be interchangeable.

eddy3 On 22 Jan 2021 - 9:11

The question as given is perfect for a technique from contest mathematics called Vieta Jumping. This is a special case of automorphism of quadratic forms. It has the virtue that it can be justified using nothing worse that the quadratic formula, and it does not require the use of square roots either. If we have a solution in positive integers to $x^2 - 5xy + y^2 = -5$ we get new ones using either $$ (x,y) \mapsto (5y-x,y) $$ or

$$ (x,y) \mapsto (x,5x -y) $$

Note that repeating one of the "jumps" twice in a row goes back to the original solution.

A "fundamental" solution is one that minimizes $x+y$ as much as possible. That is, fundamental when both $$ x+y \leq 5y - x + y $$ and $$ x+y \leq x +5x - y . $$ The first one becomes $$ 2x \leq 5 y, $$ the second $$ 2y \leq 5x .$$
Altogether, fundamental solutions are on the arc $x^2 - 5xy + y^2 $ with $$\frac{2x}{5} \leq y \leq \frac{5x}{2} $$

As you can see, there are the two integer points between the slanted lines, those being $(2,1)$ and $(1,2).$ Every solution in positive integers reduces to one of these; in turn, jumping up from one of the fundamental solutions generates all solutions.

The linear recurrence $w_{n+2} = 5 w_{n+1} - w_n$ that applies to every other number in a sequence comes from the trace and determinant of $$ \left( \begin{array}{r} 5 & -1 \\ 1 & 0 \\ \end{array} \right) $$ being $5$ and $1.$

**eddy3** · Accepted Answer

This sort of thing has a clear description in terms of Conway's topograph; I find it more convenient to use equivalent form $u^2 + uv - 5 v^2.$ The outcome for the original problem is sequences (note that the rule deals with every other element). For instance, $5\cdot 9 - 2 = 43$ and $5 \cdot 14 -3 = 67$

$$ x_{n+4}= 5 x_{n+2} - x_n,$$ $$ y_{n+4}= 5 y_{n+2} - y_n,$$

There are two interleaved subsequences.

$$ \begin{array}{c} 1&1&2&3&9&14&43&67&206&321&987 &1538&4729& \ldots \\ 3&2&1&1&2&3&9&14&43&67&206&321&987&\ldots \\ \end{array} $$ Running the indices backwards leads to different solutions, but they are just transpositions of the ones above.

Let's see, given $u^2 + uv - 5 v^2 = -5$ and $x=u+3v, y=v$ gives $x^2 - 5 xy + y^2 = -5$ and the reverse holds as well

Apparently I drew one of these in 2016

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