An Interesting Equation II

This will be another post analyzing a particular diophantine equation. Like the last time, this equation will be quadratic. Unlike last time, this time there will actually be some integral solutions. The equation in question is

where, of course, $x,y\in\Z$. In other words, we want to know which integers $y$ make $5y^2+4$ a perfect square. The answer may surprise you.

Note: I’m gonna use some facts about Dedekind domains that I haven’t proved anywhere on this blog (i.e. unique factorization of ideals into prime ideals). If I’m going to keep doing posts like this one, at some point I will write a post introducing the theory of Dedekind domains and proving that rings of integers are always Dedekind.

The Answer

The first thing to realize is that working in $K=\Q(\sqrt 5)$ allows us to rewrite our equation as

where $\knorm:K\to\Q$ is the standard relative norm, so we are looking for elements of $K$ (integral coefficients and) with norm $4$. Since we only want elements with integer coefficients, we may be tempted to work in the ring $\Z[\sqrt 5]$. However, this ring isn’t super nice. Instead, the proper setting for this problem is the ring of integers $\ints K=\Z[\phi]$ where $\phi$’s satisfies $x^2-x-1=0$ (notice that $-\inv\phi$ also satisfies this polynomial).

Hence, our goal is to find elements of $\ints K$ with norm $4$. Well, technically our goal is to find elements of $\Z[\sqrt5]$ with norm $4$, but since $\Z[\sqrt 5]\subseteq\ints K$, our only concern is that $\ints K\sm\Z[\sqrt 5]$ might have some norm $4$ elements, but this will turn out to be a non-issue.

Fix $x,y\in\Z$ such that $x+y\phi$ has norm $4$. That is,

This shows that $x+y\phi$ is a factor of $4$, so the natural thing to do would be to invoke unique factorization somehow. We are in luck because $\ints K$ turns out to be a UFD. However, this is non-trivial to prove, so we won’t rely on this fact. We’ll instead rely on the slightly-easier-to-prove facts 1 about general Dedekind domains. In particular, that they recover unique factorizations on ideals. In terms of ideals, we have

Note that $(2)$ is prime as

This makes $(2)^2$ the (unique) factorization of $(4)$ into prime ideals. Because $x+y\phi$ and $x-y\inv\phi$ are not units (as they have non-unit norms), this means we must have

Thus, $x+y\phi=2u$ for some unit $u\in\units{\ints K}$ 2. Now, we have previously shown that $\units{\ints K}=\pm\eps^{\Z}$ for some fundamental unit $\eps\in\ints K$. It’s not too hard to show furthermore that one can take $\eps=\phi$ as such a fundamental unit. This means we have shown that every norm $4$ element of $\ints K$ is of the form $\pm2\phi^n$ for some $n\in\Z$.

We can actually say more. Because $\knorm(x+y\phi)=\knorm(2)$, we must have $\knorm(u)=1$; however, $\knorm(\phi)=-1$, so $u$ must be an even power of $\phi$. That is, the norm $4$ elements of $\ints K$ are exactly those of the form $\pm2\phi^{2n}$ for some $n\in\Z$.

Since we are looking for solutions to $x^2-5y^2=4$ and squaring erases signs, we’ll ignore the $\pm$, and define sequences $a_n,b_n\in\Z$ by $a_n+b_n\phi=2\phi^{2n}$. Using that $\phi^2=1+\phi$, we see that

so $a_{n+1}=a_n+b_n$ and $b_{n+1}=a_n+2b_n$. At this point, it is worthwhile to remark that $a_0=2$ and $b_0=0$, so induction shows that $a_n,b_n$ are both even for all $n$. This means that $a_n+b_n\phi\in2\ints K\subseteq\Z[\sqrt5]$, so the norm $4$ elements are really in 1-1 correspondence with the solutions we seek.

Now, we want “integers $y$ such that $5y^2+4$ is perfect” so we really only care about $b_n$. With this in mind, note that $a_n=b_{n+1}-2b_n$ so $a_{n+1}=b_{n+2}-2b_{n+1}$. Thus, $b_{n+2}-2b_{n+1}=b_{n+1}-b_n$ which we rearrange to read

We could get an explicit formula for this 3, but we’ll cheat a little instead. If you form a table of the first few values of $b_n$,

and stare at them for long enough, you may notice that it looks a lot like $b_n=2F_{2n}$ where $F_n$ is the $n$th Fibonacci number. Indeed, defining $\ast b_n:=2F_{2n}$ we see this sequence satisfies

Since we also have $\ast b_0=2F_0=0$ and $\ast b_1=2F_2=2$, this shows that $b_n=\ast b_n=2F_{2n}$. Returning to our original question (and noting that $\phi=(1+\sqrt5)/2$), we can write $a_n+b_n\phi=x_n+y_n\sqrt5$ where

Thus, the (positive) integers $y$ such that $5y^2+4$ exactly make up every other Fibonacci number! This is a very surprising fact on first glance, but it came fairly naturally from our analysis of the equation $x^2-5y^2=4$. The fibonacci numbers showed up since we worked in the ring $\ints K=\Z[\phi]$ where $\phi$ is the golden ratio, and it’s every other fibonacci number because $\knorm(\phi)=-1$ so you switch between $5y^2+4$ being a square and $5y^2-4$ being a square as you consider consecutive powers of $\phi$ (consecutive Fibonacci numbers).

  1. that I’m still not proving here 

  2. We could have reached this conclusion earlier if we had used unique factorization at the level of elements, but the detour through ideals was a short one, so no biggy 

  3. using e.g. generating functions or linear algebra 

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