-6

I stumbled upon Fermat's theorem while reading Steig Larsson's "The Girl Who Played With Fire":

$x^3 = y^3 + z^3 \tag{Eq. 1}$

As far as it goes, it is an academic tweak of the Pythagorean theorem.

$x^2 = y^2 + z^2\tag{Eq. 2}$

So differentiating Eq. $1$ with respect to $x$, we get

$$ x^2 = (y^2)\frac{\mathrm dy}{\mathrm dx} + (z^2)\frac{\mathrm dz}{\mathrm dx}\tag{Eq. 3}$$

Now LHS of Eq. $2$ and $3$ are equal.

Eq. $3$ has scalars of $\mathrm dy/\mathrm dx$ and $\mathrm dz/\mathrm dx$ which are slopes.

Now for the RHS of Eq. $2$ and $3$ to be equal, we would want that both $y$ slope and $z$ slope be unity or $1$.

But if two sides have the same slope of $1$, then they are parallel.

But it is a given that in a right angled triangle or for that sake, any triangle cannot have parallel lines. So it is impossible for RHS of Eq. $2$ and $3$ to be same.

Does this provide a proof for Fermat's claim?

P.S Bare my follies if my question turns out to be a very stupid one. Was just curious.

Some Guy
  • 2,687
Amit Joki
  • 91
  • 4
    This question is unreadable IMO. – barak manos Jan 25 '17 at 15:24
  • @LeonMeier The OP does say "Someone please raise the powers" so I think that it's clear they mean $x^3$, etc. Overall, I don't find this question unclear - they're asking whether their argument proves FLT. What is unclear (completely, to me at least) is their argument itself, at least partly because it contains several errors (see my answer); but I'm not actually sure that means the question itself is unclear. I think it admits a potentially helpful answer, so - tentatively - I suggest it remain open. – Noah Schweber Jan 25 '17 at 15:53
  • Maybe do you would to like to read this Granville and Tucker, It's As Easy As abc, NOTICES OF THE AMS Vol. 49, No.10 (2002). On the other hand if you are interested today or in next future in nuber theory, you can see the video lectures of Murty, search in YouTube Murty, An introduction to Analytic Number Theory, from the official channel matsciencechannel. I don't know what's your background in mathematics, but with effort you can learn it, you've 40k in one of this sites, it is saying that your are very intelligent. Good luck. –  Jan 25 '17 at 15:56
  • Leon, yeah. The android app sucks and I didn't know the notations to be used. Noah got my question right. I also had specifically asked for the terms to be raised to their powers for I couldn't myself do it. On the whole it was a pretty stupid question to have been asked but happens :) – Amit Joki Jan 25 '17 at 15:58
  • No, this has nothing to do with FLT. – egreg Jan 25 '17 at 16:08
  • I edited as requested. It's still nonsense. This is what you get when you give the same name to different things and then call them equal. – DanielWainfleet Jan 25 '17 at 16:12
  • Thanks for the edit obviously I hadn't intended that. – Amit Joki Jan 25 '17 at 16:31
  • I believe that this reference also is nice and interesting: Mauldin, A Generalization of Fermat’s Last Theorem:The Beal Conjecture and Prize Problem, also in NOTICES of the AMS, Vol. 44, No. 11 (1997). –  Jan 25 '17 at 21:17
  • @user243301 thanks for those references. Have bookmarked them for near future use. – Amit Joki Jan 26 '17 at 10:59
  • Mathematics is only a technique, it is very difficult, but only is a technique and a language. From your country in the past, and currently there are good mathematicians interested in problems of number theory, other example now is Anitha Srinivasan, Should we try to solve the Markoff conjecture? (2013 available from her Google Site) and her literature about this unsolved problem. Since you are young, I know that with your effort and ideas you can do contributions as posts in this site MSE that improve this your previous question. Good luck. –  Jan 26 '17 at 15:00

1 Answers1

8

There are several mistakes here.

First, let's look back at the Pythagorean Theorem. This does not say "$x^2=y^2+z^2$," since that's blatantly false in general; rather, it says that if $x, y, z$ are the sides of a right triangle with $x$ being the hypoteneuse, then $x^2=y^2+z^2$.

A related fact is that there are infinitely many triples of real numbers $(x, y, z)$ satisfying $x^2=y^2+z^2$. Of course, this is trivial: take any $x>y>0$, and let $z=\sqrt{x^2-y^2}$.

Slightly more surprising is that there are infinitely many triples of integers satisfying that equation: e.g.

  • $5^2=3^2+4^2$

  • $13^2=12^2+5^2$

  • And infinitely many others. (See e.g. here.)

Now let's look at Fermat's Last Theorem. FLT is the statement that the fact above fails for powers $>2$: namely, it says

If $c>2$ is an integer, then there are no positive integers $x, y, z$ satisfying $x^c=y^c+z^c$.

You mention specifically the case $c=3$; note that this is just a special case of FLT, and was proved long before FLT.

Now the first problem with your argument essentially comes down to this: even if we had $x^3=y^3+z^3$, there is no reason to believe that $(x, y, z)$ is also a Pythagorean triple! So you can't hope to prove even the $c=3$ case of FLT by anything like what you write.

Moreover, since we're looking only at integer solutions (note that there trivially are real number solutions to $x^3=y^3+z^3$ - again, pick $x>y>0$ and let $z=\sqrt[3]{x^3-y^3}$), this is fundamentally a discrete problem - calculus isn't really going to come into play.

EDIT: Another, maybe more concrete, mistake occurs when you write

we would want that both y slope and z slope be unity or 1.

If $a+b=ca+db$, it is true that $c=d=1$ is one way for this to happen. But it's not the only way! For instance, $$2+2={1\over 2}\cdot 2+{3\over 2}\cdot 2.$$ So right there, you cannot deduce the thing you want (which, as I said above, isn't even relevant to the problem).

Noah Schweber
  • 245,398