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Reputable on-line sources agree that Leonard 'Fibonacci' proved the nonexistence of positive-integer solutions to $c^4 - b^4 = a^2$ . Yet my change to Wikipedia to reflect this was reverted. I hope to get a consensus on what seems like an interesting historical question.

(Reputable sources include 1. with "Fibonacci also proves many interesting number theory results such as: ... $x^4 - y^4$ cannot be a square."; and 2. ... which I can't quote because "I've reached my viewing limit." ... J.D.A.)

I've prepared a webpage summarizing the question:

Briefly, the issue seems to be about Leonardo's statement:

When $x > y$ ... then $x (x-y) \neq y (x+y)$ and from this it may be shown that no square number can be a congruum. For [then] ... the four factors $x$, $y$, $(x+y)$, $(x-y)$ must severally be squares which is impossible.

If he'd just added "by infinite descent" here, his "proof" would be valid, right?

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Looking at your webpage, I note the following sentence: "Since it is minimal, x and y have no common factors and (possibly excepting 2), neither do x+y and x-y. For this reason, any squared odd factor of t occurs in only one of the terms x, y, x+y, x-y, so they are each themselves squares." However, x+y and x-y could each be divisible by an odd power of 2 (e.g., x+y is divisible by 2, and x-y is divisible by 8), and thus not be squares. – Craig Oct 12 '11 at 13:43
The first part of the highlighted "quote" follows from the irrationality of square-root of two, a secret "discovery" of the Pythagorean school: $x(x-y) = y(x+y)$ implies $2x^2 = (x+y)^2$. Although a full proof appears in Euclid's Elements, Book X, Prop. 117, historians believe it to be an interpolation. However it would seem to be the sort of "infinite descent" proof by contradiction that supports the possibility of a reasoned argument by Fibonacci. – hardmath Oct 12 '11 at 13:55
Craig -- you're right; I did present that badly. I think one way around this is to note that for the squares (s-t, s, s+t) to be in lowest terms, one of (x, y) must be even, the other odd, so in fact neither x+y nor x-y is divisible by 2. (From the brief excerpt of Liber Quadratorum I have, I don't see how Leonardo argues, but this is not the part of Leonardo's "proof" that Anderson-Wilson objected to.) – James D. Allen Oct 12 '11 at 18:52
You might find of interest my speculative reconstruction of Fibonacci's Lost Theorem = FLT_4. – Bill Dubuque Oct 12 '11 at 22:51
Direct quotes and translations thereof are generally most likely to be accepted. – dfeuer Nov 28 '11 at 2:57

Sorry for posting a negative reply to this enjoyable question, but if the answer is supposed to be based upon what Leonardo wrote in his Liber quadratorum, I think it has to be no.

Leonardo of course did not consider any case of Fermat’s Last Theorem, but he did make a claim from which it is possible to deduce the $n=4$ case of FLT. The claim is that when you have three squares $x^2<y^2<z^2$ in arithmetic progression, the common difference $D=y^2-x^2=z^2-y^2$ can never be a square.

Unfortunately, Leonardo’s argument for his claim is not generally considered valid. Oystein Ore says in his Number Theory and its History that “[Leonardo] did not possess any satisfactory proof for it.” Andre Weil is more blunt in his “Number Theory: An Approach Through History from Hammurapi to Legendre”, p. 14, and says that “[Leonardo] asserts that $D$ can never be a square, giving for this a totally inadequate reason”.

So what was Leonardo’s argument? In order to be able to judge it, we first need to look at his Congruum Theorem, which shows how to find squares $x^2<y^2<z^2$ in arithmetic progression, so that $y^2-x^2=z^2-y^2=D$. Leonardo calls the number $D$ a congruous number, or congruum, because it makes the three squares congruent. His tool is to represent each square as a sum of odd integers, $$x^2=1+3+\dots+(2x-1)$$ $$y^2=1+3+\dots+(2y-1)$$ $$z^2=1+3+\dots+(2z-1)$$ We then get two different representations of the congruous number $D$, $$D=(2x+1)+\dots+(2y-1)=(2y+1)+\dots+(2z-1)$$ Searching for adjacent sets of consecutive odd numbers with the same total sum, Leonardo finds the following solutions:

Let $m$ and $n$ be positive integers with $m<n$.

(i) If $m$ and $n$ have the same parity and ${n\over m}<{n+m\over n-m}$ then a solution is given by $$x={1\over 2}(m(n+m)-n(n-m))$$ $$y={1\over 2} (m(n+m)+n(n-m))={1\over 2} (n(n+m)-m(n-m))$$ $$z={1\over 2} (n(n+m)+m(n-m))$$ $$D=nm(n+m)(n-m)$$ (ii) If $m$ and $n$ have the same parity and ${n\over m}>{n+m\over n-m}$ then a solution is given by $$x={1\over 2} (n(n-m)-m(n+m))$$ $$y={1\over 2} (n(n-m)-m(n+m)) ={1\over 2} (n(n+m)-m(n-m))$$ $$z={1\over 2} (n(n+m)+m(n-m))$$ $$D=nm(n+m)(n-m)$$ (iii) If $m$ and $n$ have opposite parity and ${n\over m}<{n+m\over n-m}$ then a solution is given by $$x=m(n+m)-n(n-m)$$ $$y=m(n+m)+n(n-m)=n(n+m)-m(n-m)$$ $$z=n(n+m)+m(n-m)$$ $$D=4nm(n+m)(n-m)$$ (iv) If $m$ and $n$ have opposite parity and ${n\over m}>{n+m\over n-m}$ then a solution is given by $$x=n(n-m)-m(n+m)$$ $$y=n(n-m)-m(n+m)=n(n+m)-m(n-m)$$ $$z=n(n+m)+m(n-m)$$ $$D=4nm(n+m)(n-m)$$ (Note that Leonardo does not say that this list contains all possible solutions, as might have been expected if he had reasoned from the classification of Pythagorean triples.)

It is then natural to consider what would happen if ${n\over m}={n+m\over n-m}$, and Leonardo proves that this can never be the case. We would do this by some simple algebra, for instance by setting $x={n\over m}$ and observing that the equation $x={x+1\over x-1}$ has no rational solution. For Leonardo it is more natural to look back at what he has already shown, and to observe that if ${n\over m}={n+m\over n-m}$ then we would get $x=0$ in all the solutions above. We would then have three squares in arithmetic progression starting with $x=0$, where $y^2$ would be equal to the congruous number $D$ and $z^2$ would be equal to $2D=2y^2$. This leads to a contradiction since $\sqrt 2$ is not a rational number.

Note in particular that Leonardo has shown that if we were to have ${n\over m}={n+m\over n-m}$ then $D$ would be a square and we would get a contradiction.

We now come the the heart of the matter. Immediately following the argument above, Leonardo says: “From this will be shown, in fact, that no square number can be a congruous number; because if it were possible, then the ratio of the sum of the two given numbers to the difference would be as the larger of them to the smaller.”

In other words, Leonardo says that if $D$ were to be a square, then we would get ${n+m\over n-m}={n\over m}$. This is all the explanation that is given, and Leonardo does not come back to this or similar questions in the book.

The original question above quotes a statement from an article from 1919 by R. B. McClenon, reprinted in the book Sherlock Holmes in Babylon and Other Tales of Mathematical History by Anderson and Wilson. Note that the phrase: “For [then] … the four factors $x$, $y$, $x+y$, $y-x$ [which are $m$, $n$, $n+m$, $n-m$ above] must severally be squares, which is impossible” does not come from Leonardo, but is the kind of addition that McClenon would like to make in order to complete the proof. There are no considerations of this type in Liber quadratorum, though.

I also don’t think Leonardo could have saved his argument by writing “by descent” next to it any more than Fermat could have written “modular elliptic curves” next to his marginal comment. There are no considerations of minimal solutions present anywhere in Liber quadratorum, and Leonardo would have needed considerable space to present such a foreign idea to his readers.

To me, the most likely hypothesis is that Leonardo made a rare mistake and erronously thought that the converse of what he had just shown would follow automatically. He would of couse be strengthened in this belief by having computed numerous examples of congruous numbers and squares in arithmetic progression.

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Another vote in the negative: Dickson writes (History of the Theory of Numbers, Volume II, page 615), "Leonardo Pisano recognized the fact, but gave an incomplete proof, that no square is a congruent number (i.e., $x^2+y^2$ and $x^2-y^2$ are not both squares)...." – Gerry Myerson Jun 26 '12 at 6:13

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