Consider the identity $(b-a)(b+a) = ab - 1$, where $a, b$ are nonnegative integers.

We can also express this identity as $a^2 + ab - b^2 = 1$.

This identity is clearly true when $a = F_{2i-1}$ and $b = F_{2i}$, where $F_i$ is the $i^{th}$ term of the Fibonacci sequence. This is equivalent to one case of Cassini's identity, $(F_{2i-1})(F_{2i+1}) = F_{2i}^2 + 1$, and is easily proven by induction or several other simple elementary means.

My question is this: Is there a simple elementary proof that these Fibonacci numbers are the only solutions of this identity?

By simple elementary proof, ideally I mean a proof using methods and steps that a mathematically gifted high school student could follow and understand. Alternatively, I could define it as a proof using methods that would have been known to mathematicians in Cassini's time, in the late 17th century. In other words, I am looking for a proof that does not rely on more advanced methods such as quadratic number fields or generalized solutions of Pell equations.

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    $\begingroup$ For the benefit of readers who are interested in the more advanced solution, the first step is to observe that $a^2 + ab - b^2$ is the norm form $N(a + b \phi)$ where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio and $N$ refers to the norm in the ring of algebraic integers $\mathcal{O}_{\mathbb{Q}(\sqrt{5})} = \mathbb{Z}[\phi]$. So we can find all solutions by computing generators for the unit group, since units are precisely the elements of norm $\pm 1$. $\endgroup$ Commented Dec 3, 2017 at 19:19
  • $\begingroup$ Very simple with Conway's Topograph diagram, see new book bookstore.ams.org/mbk-105 $\endgroup$
    – Will Jagy
    Commented Dec 3, 2017 at 19:26
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    $\begingroup$ Ok yes rtybase after careful study of the answer you provide there, I accept that it also answers my question. The key point, for me, is your argument in the "descent" step that n>q>0 or n=q=1, because if q>=n, then q^2+nq-n^2 >= n^2+n^2-n^2=n^2>=1. But q^2+nq-n^2=1 or -1, so q>=n can only be true if n=q=1. This is the key step to make the proof complete. Now we can effectively use the descent method to show that the sequence must eventually reach n=q=1, so all of the solutions must indeed be Fibonacci numbers. Thank you! $\endgroup$ Commented Dec 4, 2017 at 0:43

3 Answers 3


Here's one approach:

Step 1a: Show that if $a$ and $b$ satisfy this, and $0 < a < b$, then $a' = (b-a)$ and $b' = a$ also satisfy it and have $a' \le b' < b$, so that the max absolute value of the two items in the pair decreases

What the heck...let's check that: we want to show that $(b' - a') (b' + a') - (a'b' -1) $ is zero. So compute \begin{align} (b' - a') (b' + a') - (a'b' -1) &=(a - (b-a)) (a + (b-a)) - (a(b-a) -1)\\ &=(2a - b)) (b)) - (a(b-a) -1)\\ &=2ab - b^2 - ab + a^2 +1 \\ &=ab - b^2 + a^2 +1 \end{align} which is $0$ because $a$ and $b$ satisfy the relation, which expanded out says that $b^2 - a^2 - ab + 1 = 0$.

Case 1b: if $b < a < 0$, then $b' = b-a$ and $a' = b$ do as well, and $b' < a' < 0$, and $|b'| = |a| < |b|$. Proof: exactly the same as before. Once again, the max absolute value of the two items in the pair decreases.

Case 1c: $b$ and $a$ have opposite signs. If $b$ is positive, then $a$ is negative, and $|a| > |b|$. If $b$ is negative, then $a$ is positive, so $b-a$ is negative, so $b+a$ is positive, and once again $|a| > |b|$. Again, by an argument like the one above, the pair $(a, b)$ can be adjusted to a pair $(b, a-b)$ where the larger number (in absolute value) is smaller in the new pair than in the old one, i.e., the max absolute value of the two items in the pair decreases.

Other cases: you still have to deal with other similar cases in ways rather like this, and I don't have the stomach to go through it all.

Step 2: Conclude that for any such pair, we can reduce the pair to a smaller (in the sense of max-absolute-value) pair of numbers, until until $a = b$ (which fails unless $a = b = \pm 1$).

Step 3: Conclude that our pair is part of the sequence that arises from $(1, 1) \to (1, 2) \to (2, 3) \to (3, 5) \to ...$, i.e., the F-sequence.

[This does only handle the case where $0 < a < b$; the $a=b$ case is trivial (indeed, Step 2 addresses it); the $a > b$ case can almost certainly be handled by essentially the same method. The case where $a$ or $b$ is zero should not be difficult for a bright high-school student.]

  • $\begingroup$ A small comment is likely necessary for the $(1, 0)$ case, but otherwise I like this answer! $\endgroup$
    – B. Mehta
    Commented Dec 3, 2017 at 19:14
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    $\begingroup$ This is classical problem with Vieta jumping solution. $\endgroup$
    – nonuser
    Commented Dec 3, 2017 at 19:16
  • $\begingroup$ Thanks, @B.Mehta; I'd forgotten to mention those cases. John Watson: I guess I need to read about Vieta jumping solutions -- is that the general case of what I just wrote out? $\endgroup$ Commented Dec 3, 2017 at 19:17
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    $\begingroup$ en.wikipedia.org/wiki/Vieta_jumping $\endgroup$
    – nonuser
    Commented Dec 3, 2017 at 19:19
  • $\begingroup$ See my comment on the other answer below. The difficulty is that the identity can also be valid for negative integers as well, so hypothetically it is possible that such a sequence could go from large positive numbers to negative numbers in one step, and continue with smaller negative numbers (larger absolute value) from there to negative infinity. A proof must rule out this possibility. $\endgroup$ Commented Dec 3, 2017 at 19:21

Open this up to solving $a^2+ab-b^2=\pm1$ in positive integers. Unless $a$ and $b$ are very small, then $a<b$. Let $c=b-a$. Then $$c^2+ca-a^2=(b-a)^2+(b-a)a-a^2=-a^2-ab+b^2=\mp1.$$ So if $c$ and $a$ are consecutive Fibonacci's then so are $a$ and $b$.

To complete this, one needs to analyse the solutions for small $a$ and $b$ in order to start off the induction.

  • $\begingroup$ Thank you for the replies and ideas, but the difficulty with this approach of proof by "descent" to smaller terms is that the identity can also be valid for negative integers as well, which can also obey Fibonacci-like sequence relations. So one cannot say that an arbitrary a, b will necessarily reduce to small positive numbers like a=b=1. Hypothetically it could be possible that such a sequence could go from large positive numbers to negative numbers in one step, and continue with smaller negative numbers (larger absolute value) from there to negative infinity. A proof must rule this out. $\endgroup$ Commented Dec 3, 2017 at 19:18

It turns out I have a diagram of Conway's topograph pdf for this. The very simple statement is that "the river is periodic." This means that, if we can find all solutions within one period, we have them all.

Recent book by Allen Hatcher pdf

ALSO: Recent book at a fairly elementary level: Weissman

Broken down further, it means that any solution to $a^2 + ab - b^2 = 1$ leads to another, $$ (a,b) \mapsto (a+b, a + 2b) $$ As you can see from the (vertical) vectors when the form value is $1,$ this makes $a,b$ consecutive Fibonacci, by induction.

As you can see, I draw in little $(x,y)$ "coordinate" pairs as column vectors. This is crucial for my approach; the other two books do not really push this aspect, but it is done well in Stillwell, Elements of Number Theory.

enter image description here

I have drawn a portion of the river, with colors as in the tree diagrams, and showing the relative positions of the values $11.$ I have worked out how to force the given mapping $ (a,b) \mapsto (a+b, a + 2b) $ as we move to the right, or $ (a,b) \mapsto (2a-b, -a + b) $ as we move to the left.

enter image description here

There was a question in comment about $a^2 + ab - b^2 = 11.$ It is enough to draw a single "tree" of positive values climbing away from the river. We see $11$ as $(a,b)$ pairs $(3,1)$ and $(3,2).$ All other solutions with positive $(a,b)$ occur in other trees along the river. They can be found with $ (a,b) \mapsto (a+b, a + 2b) .$ In the next tree to the right, we get $(4,5)$ and $(5,7).$ A second tree to the right, we get $(9,14)$ and $(12,19).$ Also, Cayley-Hamilton says that we get two orbits under a pair of linear degree two recurrences, $$ a_{n+2} = 3 a_{n+1} - a_n, $$ $$ b_{n+2} = 3 b_{n+1} - b_n. $$ I wrote out a simple proof without using Cayley-Hamilton at How does one solve this recurrence relation?

enter image description here

Alright, I did one tree over, I mostly left off the blue edge labels, which match the previous tree.

enter image description here

Other answers/questions I did with the topograph:













http://math.stackexchange.com/questions/1132187/solve-the-following-equation-for-x-and-y/1132347#1132347 <1,-1,-1>









  • $\begingroup$ Thank you for the illustration, but I just wonder, doesn't the math behind Conway's topograph require essentially the same advanced abstract algebra methods that Qiaochu Yuan mentioned in his answer above? I don't believe that an automorphism group is a concept that a high school student or a 17th century mathematician could comprehend. $\endgroup$ Commented Dec 3, 2017 at 19:47
  • $\begingroup$ Also, are you sure that this method proves the impossibility of other solutions in other sequences? For example, let's say I claimed all solutions of (b-a)(b+a) = ab - 11 were in the sequence 1,4,5,9,14,23,37,60,.... As in your method, induction shows that each solution in the sequence leads to another. But in fact in this case there is a different sequence with solutions: 2,5,7,12,19,31,50,.... Can your method identify both sequences? How? $\endgroup$ Commented Dec 3, 2017 at 20:05
  • $\begingroup$ @GeoffreyCaveney yes, I'm sure. When you switch to a prime, in your case $11,$ you expect two orbits. Forget the word automorphism. It is just an explicit integer mapping that takes locations in the topograph to others with the same quadratic form. Recommend you learn the method, here is Conway's original maths.ed.ac.uk/~aar/papers/conwaysens.pdf $\endgroup$
    – Will Jagy
    Commented Dec 3, 2017 at 20:13
  • $\begingroup$ @GeoffreyCaveney added links to some topograph answers. Recommend you learn how to draw them. Also, given your concerns, suggest you look at math.stackexchange.com/questions/739752/… where, because $55$ is composite, there are several orbits; these are readily found in the "trees" I draw. $\endgroup$
    – Will Jagy
    Commented Dec 3, 2017 at 20:19
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    $\begingroup$ Just to clarify for other readers, Conway's concepts like "the river is periodic" that Will mentions do make sense, but only if you carefully work through the First Lecture of Conway's book The Sensual (Quadratic) Form, "Can You See The Values of 3x^2 + 6xy - 5y^2?" The terms "topograph", "river", "lake", "tree", "well", etc., have specific definitions in the context of Conway's treatment of quadratic forms, lattices, matrices, vectors, bases, norms, etc. One must work through all of this material, for terms like "the river" to make mathematical sense. $\endgroup$ Commented Dec 3, 2017 at 20:45

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