| ... I think Pepin is being unfair to Euler when he writes
|
| Next, having recognized that the only solution for which
| $y$ equals +1 (or -1) corresponds to the values p=q=+1/2
| (or both -1/2), that render x=+2 (or -2), y=+1 (or -1),
| Euler concludes that 4 is the only square which answers
| the question. [*]
|
| Do you agree?
Well, let us first see what Euler says to this respect. Pepin [*]
is referring to Euler's *Article 194* ("Algebra", II, chapter 12;
or, Opera Omnia, (1), I, 432-433), where Euler is looking for perfect
squares (x^2) such that $5xx+7$ can be a cube.
Euler's way of dealing with $5xx+7=z^3$ was first to consider it as an
instance of $axx+cyy=z^3$, so that he could use *Article 188*, with
$a=5$ and $c=7$, which implies the two real equations $x=5p^3-21pqq$
and $y=15ppq-7q^3$ for some as-yet-undetermined numbers $p$, $q$.
We also know that $y=\pm 1$, and so the real equations become
$15ppq-7q^3=q(15pp-7qq)=\pm 1$. Therefore $q$ must be a divisor of 1.
Hence $q=\pm 1$, and $15pp-7=\pm 1$. In both cases, $p$ gives irrational
numbers; from this, however, we cannot infer that this question is
not at all possible, because $p$ and $q$ could be fractions such that
$x$ be a whole number when $y=1$. Fortunately, this occurs when $p=1/2$
and $q=1/2$, and so, $x=2$, $y=1$. With other fractions, however, the
'issue' is not possible.
============
194
Frage III: Man verlangt solche fuenffache Quadrate, wann
dazu 7 addirt wird dass ein Cubus herauskomme: oder dass
$5xx+7$ ein Cubus sey?
Man suche erstlich diejenigen Faelle da $5xx+7yy$ ein
Cubus wird, welches nach dem Articul 188, wo $a=5$ und
$c=7$, geschieht, wann $x=5p^3-21pqq$ und $y=15ppq-7q^3$;
weil nun hier seyn soll $y=\pm 1$, so wird
$15ppq - 7q^3 = q(15pp - 7qq) = \pm 1$
da dann $q$ ein Theiler seyn muss von 1, folglich $q=1$;
daher wird $15pp-7=\pm 1$, wo beyde Faelle fuer $p$ etwas
irrationales geben, woraus aber doch nicht geschlossen
werden kann, dass diese Frage gar nicht moeglich sey,
weil $p$ und $q$ solche Brueche seyn koennten, da $y=1$
und $x$ doch eine gantze Zahl wuerde; solches geschieht
wuercklich wann $p=1/2$ und $q=1/2$, dann da wird $y=1$
und $x=2$, mit andern Bruechen aber ist die Sache nicht
moeglich.
Pepin, of course, is correct to say that "Euler concludes that 4 is
the only square which answers the question" (since _mit andern Bruechen
aber ist die Sache nicht moeglich_). Yet, he is being somehow unfair
to Euler.
Why? ...
Because I wholeheartedly agree with your previous remarks, when you
suggest that Euler, in Chapter XII, is not offering his analysis as
any kind of *rigorous* proofs. His *Articles* (!) are more in the
spirit, as you put it, of "Here is an ingenious way of generating
solutions to problems of this type", but mind that "Sometimes there
are solutions that do not arise from this method, and it would be
an important issue to *understand* where they come from".
We should remember also that a lot of Euler's work on this topic was
*exploratory*, and of course that it was all done in advance of the
"complete" [whatever that might mean] proofs achieved in the second
half of the XIXth century.
| Also, in view of
|
| 3. We cannot regard any of the two Fermat's theorems as entirely
| proven, which is the object of the first two questions of Chapter
| XII, as long as we have not justified the use of the preceding
| method for the two formulae x^2 + y^2, x^2 + 2y^2. But, this is
| what was neither worked out by Euler nor by Legendre. On this issue,
| Legendre contented himself with quoting Euler's demonstrations.
|
| it would be interesting to know what Legendre said, and whether he
| treated Euler's partial analysis as if it were (or claimed to be) a
| full solution.
According to Dickson, "A.M. Legendre treated Fermat's problems as had
Euler". The reference for Legendre is his classic "The/orie des nombres",
Article 336, p.12, II, 3rd ed., 1830. Unfortunately, the book was not on
shelf; so I could not check what Legendre *actually* said. Any kind soul
out there willing to help? ...
|
| Thanks, Julio, for your industrious work on translating these passages!
| I hope others are benefitting from them as much as I have.
You're most welcome!
| Jim Propp
| Department of Mathematics
| University of Wisconsin
With best wishes from Montevideo,
Julio Gonzalez Cabillon
PS For the friend that today privately wrote to me "I've been trying for
the last week to respond to your kind greeting for 1999, but I couldn't
get your email address (jgc@adinet.com.uy) to work", as well as others
that might have had similar problems with ADINET.COM.UY (despues de tanto
tiempo... todavia continuan los problemas, queridos amigos de ADINET?),
I suggest you to try the *invincible* jgc@chasque.apc.org (if this email
address also fails, then you may conclude that URUGUAY is out of the Net!