## Golden numbers are Fibonacci

This post is fourth in a series, proving the curious fact that $n$ is a Fibonacci number if and only if one (or both) of $5n^2 + 4$ or $5n^2 - 4$ is a perfect square; we call numbers of this form golden numbers. Last time, I presented Gessel’s proof that every Fibonacci number is golden. In this post, I will present Phillip James’s proof that every golden number is Fibonacci.

First, we need the following lemma:

Lemma. If $x$ and $y$ are positive integers such that

$y^2 - xy - x^2 = \pm 1$ (**),

then $(x,y) = (F_n, F_{n+1})$ for some $n \geq 1$.

Let’s save the proof of this until later, noting only that the above equation will come up often, so we give it the name (**) to be able to refer to it easily. First, let’s see how we can use this lemma to prove the main result.

Theorem. If $5x^2 \pm 4$ is a perfect square, then it is a Fibonacci number.

Proof. Consider equation (**) again. We can think of this as a quadratic equation in $y$, writing it as

$y^2 + (-x)y + (-x^2 \pm 1) = 0$

for emphasis. We can now use the quadratic formula to solve for $y$, obtaining

$y = \frac{1}{2}(x + \sqrt{x^2 - 4(-x^2 \pm 1)}) = \frac{1}{2}(x + \sqrt{5x^2 \pm 4}).$

(We want only the positive solution for $y$.) We now claim that if $5x^2 \pm 4$ is a perfect square, then $y$ is an integer. If $5x^2 \pm 4$ is a perfect square, then it is easy to see that $x + \sqrt{5x^2 \pm 4}$ is an integer; the only potential problem is that factor of $1/2$. We can see that $y$ will be an integer exactly when $x + \sqrt{5x^2 \pm 4}$ is even. First, note that $\sqrt{5x^2 \pm 4}$ will have the same parity (evenness or oddness) as $x$: squaring, multiplying by an odd number, adding an even number, and taking the square root of a perfect square are all operations that preserve parity. The final expression $x + \sqrt{5x^2 \pm 4}$ then involves adding (or subtracting) two integers of the same parity, that is, both even or both odd. This will always result in an even integer.

So, if $x$ is a positive integer such that $5x^2 + 4$ is a perfect square, then $y$ as defined by the above equation will also be a positive integer, such that equation (**) holds. By Lemma 2 above, this means $x$ (and $y$) must be Fibonacci numbers. QED.

Proof of Lemma. The only thing we have left is to prove the lemma: if $x$ and $y$ are positive integers such that $y^2 - xy - x^2 = \pm 1$, then $x$ and $y$ are consecutive Fibonacci numbers.

The proof is by (strong) induction on the sum $x + y$. Strong induction means that in the inductive step, we get to assume that the lemma holds for all $x$ and $y$ with a smaller sum. With normal (weak) induction, when proving a case with $x + y = n$ we would only get to assume the lemma for the case $x + y = n-1$. It turns out that strong and weak induction are logically equivalent, even though strong induction seems, well, stronger. Perhaps that’s a post for another time!

We consider three cases: either $x > y$, $x = y$, or $x < y$.

• The case $x > y$ is actually impossible, since we would have

$y^2 - xy - x^2 < y^2 - y^2 - y^2 = -y^2$

But since $y$ is a positive integer, $-y^2 \leq -1$, and so we would have $y^2 - xy - x^2 < -1$, which is impossible since we assumed it is equal to $\pm 1$.

• Next, if $x = y$, then $y^2 - xy - x^2 = -y^2 = \pm 1$, so we must have $y = 1$ and hence $x = 1$ as well. These are consecutive Fibonacci numbers (namely, $F_1$ and $F_2$) so the lemma holds in this case.

• The last case is $x < y$. Consider $(a,b) = (y-x, x)$, which by assumption are both positive integers. It turns out that they also satisfy (**):

$\begin{array}{rcl}b^2 - ba - a^2 & = & x^2 - x(y-x) - (y-x)^2\\ &=& x^2 - xy + x^2 - y^2 + 2yx - x^2\\ &=& x^2 + xy - y^2\\ &=& -(y^2 - xy - x^2)\\ &=& \pm 1\end{array}$

Since $a + b = y-x + x = y < x+y$, by the inductive hypothesis we conclude that $a$ and $b$ are consecutive Fibonacci numbers, say $(a,b) = (F_n, F_{n+1})$. But then $(x,y) = (b,a+b) = (F_{n+1}, F_{n+2})$ are also consecutive Fibonacci numbers.

And that concludes the proof that all golden numbers are Fibonacci. I note that James also proves the converse of the Lemma (that is, he proves that any consecutive Fibonacci numbers satisfy equation (**)), and uses it to prove that all Fibonacci numbers are golden; you might enjoy trying to complete the proof yourself using some of the above ideas.

I’ll conclude this series with a post exploring the question of how quickly we can test an arbitrary integer for Fibonacci-ness, and in particular whether this theorem might give us a faster test than other methods. But first—Carnival of Mathematics #132 will be up soon!

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 2 Responses to Golden numbers are Fibonacci

1. Justin says:

Is there a glossary somewhere for math terms that you use? I almost always need to start by reminding myself what a ‘lemma’ is, and it would probably be useful to have a more specific understanding of what things like theorems are.

• Brent says:

Not really, but I’m happy to answer questions! A theorem is a mathematical statement that has been proven, i.e. it has been logically shown why it must be true. There’s really no difference between a lemma and a theorem, other than the fact that calling something a ‘lemma’ usually indicates that we are not especially interested in it per se, but intend to use it to prove something else we do care about. In other words, a lemma is a sort of “helper theorem”.