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I am trying to understand the proof in Carmichaels book Diophantine Analysis but I have got stuck at one point in the proof where $w_1$ and $w_2$ are introduced.

The theorem it is proving is that the system of diophantine equations:

  • $$x^2 + y^2 = z^2$$
  • $$y^2 + z^2 = t^2$$

cannot simultaneously be satisfied.


The system is algebraically seen equivalent to

  • $$t^2 + x^2 = 2z^2$$
  • $$t^2 - x^2 = 2y^2$$

and this is what will be worked on. We are just considering the case where the numbers are pairwise relatively prime. That implies that $t,x$ are both odd (they cannot be both even). Furthermore $t > x$ so define $t = x + 2 \alpha$.

Clearly the first equation $(x + 2\alpha)^2 + x^2 = 2 z^2$ is equivalent to $(x + \alpha)^2 + \alpha^2 = z^2$ so by the characterization of primitive Pythagorean triples there exist relatively prime $m,n$ such that $$\{x+\alpha,\alpha\} = \{2mn,m^2-n^2\}.$$

Now the second equation $t^2 - x^2 = 4 \alpha (x + \alpha) = 8 m n (m^2 - n^2) = 2 y^2$ tells us that $y^2 = 2^2 m n (m^2 - n^2)$ by coprimality and unique factorization it follows that each of those terms are squares so define $u^2 = m$, $v^2 = n$ and $w^2 = m^2 - n^2 = (u^2 - v^2)(u^2 + v^2)$.


It is now said that from the previous equation either

  • $u^2 + v^2 = 2 {w_1}^2$, $u^2 - v^2 = 2 {w_2}^2$

or

  • $u^2 + v^2 = w_1^2$, $u^2 - v^2 = w_2^2$

but $w_1$ and $w_2$ have not been defined and I cannot figure out what they are supposed to be. Any ideas what this last part could mean? Thanks a lot!


for completeness, if the first case occurs we have our descent and if the second case occurs $w_1^2 + w_2^2 = 2 u^2$, $w_1^2 - w_2^2 = 2 v^2$ gives the descent. Which finishes the proof.

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2 Answers 2

up vote 10 down vote accepted

This descent has a very beautiful presentation based upon ideas going back to Fibonacci.

Fibonacci's Lost Theorem $\: $ The area of an integral pythagorean triangle is not a square integer.

Over 400 years before Fermat's celebrated proof by infinite descent of the essentially equivalent $\rm\:FLT_4\:$ (Fermat's Last theorem for exponent $4$), Fibonacci claimed to have a proof of this in his Liber Quadratorum (Book of Squares). But, alas, to this day, his proof has never been found.

Below is my speculative reconstruction of Fibonacci's proof of this theorem, based upon similar ideas that survived in his extensive studies on squares and related topics.

A square arithmetic progression (SAP) is an AP $\rm\ x^2,\ y^2,\ z^2\ $ with a square stepsize $\rm\ s^2\:,\ $ viz. $$\rm\ x^2\ \ \xrightarrow{s^2}\ \ y^2\ \ \xrightarrow{s^2}\ \ z^2$$

Naturally associated with every SAP is a "half square triangle", ie. doubling $\rm\ z^2 + x^2\ $ produces a triangle of square area $\rm\ s^2\ $ viz. $\rm\ (z + x)^2 + (z - x)^2\ =\ 2\ (z^2 + x^2)\ =\ 4\ y^2\ $

which indeed has $\ \ $ area $\rm\ =\ (z + x)\ (z - x)/2\ = \ (z^2 - x^2)/2\ =\ s^2\ $

With these concepts in mind, the proof is very easy: If there exists a pythagorean triangle with square area then it may be primitivized and its area remains square. Let its primitive parametrization be $\rm\:(a,b)\:$ and let its area be $\rm\:c^2\:,\:$ namely

$$\rm\ \frac{1}2\ leg_1\ leg_2\ =\ \frac{1}2\ (2\:a\:b)\ (a^2-b^2)\ =\ (a-b)\ a\ (a+b)\ b\ =\ c^2\:.\ $$

Since $\rm\:a\:$ and $\rm\:b\:$ are coprime of opposite parity, $\rm\ a-b,\ a,\ a+b,\ b\ $ are coprime factors of a square, hence must all be squares. Then $\rm\ a-b,\ a,\ a+b\ $ form a SAP; doubling its half square triangle yields a triangle with smaller square area $\rm\ b < c^2\:,\ $ hence descent. $\ \ $ QED

Note: the doubling construction was already in Euclid; it may be viewed as a composition of forms $\rm\ (z^2 + x^2)\ (1^2 + 1^2)\:. $

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1  
This is the most beautiful proof I have seen in ages.. thanks for posting that! –  quanta Mar 16 '11 at 2:05
    
Thanks, I'm happy you enjoyed it. There is also a pretty diagram that illustrates the descent but, alas, it doesn't fit in the margin. More seriously, it was among some of my items that were stolen while in storage, and I've never had the time to reconstruct it. But it shouldn't be too difficult. –  Bill Dubuque Mar 16 '11 at 2:17
    
Van der Poorten gives a lovely descent proof (maths.mq.edu.au/~alf/SomeRecentPapers/183.pdf) of the fact that no four squares are in AP. It would be interesting to compare your (remarkable) proof with his, to see what they have in common. –  Kieren MacMillan Oct 4 '13 at 20:58

$u^2$ and $v^2$ are $m$ and $n$, respectively, which are coprime. Then since $(u^2+v^2)+(u^2-v^2)=2u^2$ and $(u^2+v^2)-(u^2-v^2)=2v^2$, the only factor that $u^2+v^2$ and $u^2-v^2$ can have in common is a single factor of $2$. Since their product is the square $w^2$, that leaves the two possibilities given.

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