**Fermat’s Lost Theorem:** Show that $latex (x+y)^n = x^m+y^m$ has only one solution for integers $latex x > y > 0$ and $latex m,n > 1$.

The truth is, I don’t know how to solve this problem myself. But, I think that we can figure it out together. Below, I will give the solution to the simpler case of $latex y=1$ and $latex x > 1$. I expect that many of you know the following variant of the problem:

**Fermat’s Last Theorem: **Show that $latex z^m = x^m + y^m$ has no solutions for integers $latex z > x > y > 0$ and $latex m > 2$.

The above theorem was one of the most important unsolved problems in mathematics, until Andrew Wiles presented his proof to the public at a conference in Cambridge in 1993. Then someone pointed out a serious flaw in his proof and the extreme high Wiles was feeling turned into a dark abyss of despair. But being awesome implies that you pick yourself up and run full force towards the wall as if you didn’t get floored the last time you tried breaking through. And so Andy went back to his office and, with a little help from his friend (and former student) Richard Taylor, he fixed the flaw and published the massive proof in the most prestigious journal of mathematics, the Annals of Mathematics, in 1995.

But, why did Fermat’s Last Theorem become so famous in the first place? The story goes that Pierre de Fermat (a lawyer and amateur skydiver) wrote the following innocuous looking sentence in his copy of Diophantus‘ Arithmetica: “*I have discovered a truly remarkable proof which this margin is too small to contain.*” Yeah, right. And I am the Duke of Yorkshire. Still, Fermat had proved some pretty cool theorems already, so many mathematicians took his claim seriously. That was back in 1637… The proof of Fermat’s Last Theorem took 357 years to complete and, as Fermat mentioned correctly, it could not fit in the margins of a book: The published proof is over 100 pages long.

But enough about Fermat’s Last Theorem. You are here for Fermat’s *Lost Theorem*. Legend has it that Fermat came up with the solution to the special case of his Last Theorem, where $latex z=(x+y)^p$ and $latex m=ncdot p$. The page containing the solution to the special case was burned at the great fire of Toulouse in 1463 (it was still burning 170 years later…) and so the legend was born. It was not until 1993, the same year Wiles presented the flawed proof of Fermat’s *Last Theorem*, that a relatively unknown Romanian mathematician came across the *Lost Theorem* and decided to give it to Romanian High School students to solve (check out the Romanian Math Olympiad of 1993). Below, I will try to reconstruct the (alleged) solution of Fermat to the following special case:

**Fermat’s Little Lost Theorem: **Show that $latex (x+1)^n=x^m+1^m$ has a unique solution for $latex x > 1$ and $latex m,n > 1$.

**Solution: **We will need the following important equation which holds for any number $latex x$ and integer $latex n$: $latex x^n-1 = (x-1)(x^{n-1}+x^{n-2}+ldots+x^2+x+1)$. The proof of this identity is simple: Multiply each term in the second parenthesis with $latex x$ to get $latex x^n+x^{n-1}+x^{n-2}+ldots+x^2+x$ and subtract $latex x^{n-1}+x^{n-2}+ldots+x^2+x+1$ to get $latex x^n-1$ (since every other term cancels in the telescoping sum!) We have shown that the fraction $latex frac{x^n-1}{x-1}$ is always equal to $latex x^{n-1}+x^{n-2}+ldots+x^2+x+1$. In fact, you have just seen the formula for a **geometric series**, which finally answers *Problem 2* in Deal or no deal? But we are not done exploiting this formula: Since

$latex frac{x^n-1}{x-1} = x^{n-1}+x^{n-2}+ldots+x^2+x+1,$

this implies that:

$latex x^{n-1}+x^{n-2}+ldots+x+1 = (x^{n-1}-1)+(x^{n-2}-1)+ldots+(x-1)+(1-1)+n = (x-1)cdot M + n,$

(we subtracted $latex n$ ones and added $latex n$ at the end to balance the equation), where $latex M = (x^{n-2}+x^{n-3}+ldots+x+1)+(x^{n-3}+x^{n-4}+ldots+x+1)+ldots+(x+1)+1$ is an integer for every integer $latex x$. Putting everything together, we see that for any integers $latex x,n$ there is an integer $latex M$ such that:

$latex (x+1)^n-1 = x^2cdot M + xcdot n,$

where we changed $latex x$ into $latex x+1$. Why is this important for solving the problem at hand? Well, $latex (x+1)^n-1$ is supposed to be equal to $latex x^m$, for some $latex m ge 2$. But, we have just shown that $latex x^m = (x+1)^n-1 = x^2cdot M + xcdot n$. In other words, $latex n = x^{m-1}+Mcdot x = x(x^{m-2}+M)$, which implies that $latex x$ must divide perfectly the power $latex n$!

We are now ready to finish the job. Let $latex x=2k+1$ be an odd integer. Then, $latex (x+1)^m = 2^m(k+1)^m$ is divisible by 4 (since $latex mge 2$). On the other hand, $latex (2k+1)^m+1$ always has remainder 2 when divided by 4 (why?) So, $latex x$ cannot be an odd number and we are left with $latex x=2^scdot r$, for $latex sge 1$ and $latex r$ an odd integer. This makes our equation look like this:

$latex (2^scdot r+1)^{n} -1 = 2^{sm}cdot r^{m}$. Let us look at the two smallest values of $latex n$, recalling that $latex x$ divides $latex n$: The smallest value is $latex n=2$, which implies $latex x=2$ and $latex m=3$. The next value is $latex n=4$, which gives: $latex (2^sr+1)^4-1 = 2^{4s}r^4+4cdot2^{3s}r^3+6cdot 2^{2s}r^2+4cdot 2^sr$. The crucial step comes next: Factor out $latex 2^{s+2}r$ from the expression on the right to get: $latex (2^scdot r+1)^{4} -1 = (2^{s+2} r) (2^{3s-2}r^3+2^{2s} r^2+3cdot 2^{s-1} r+1) = 2^{sm}cdot r^m$. Now what? Well, we are done! For the last two expressions to be equal, we must have that $latex r = 1$ and $latex s=1$, otherwise $latex 2^{3s-2}r^3+2^{2s} r^2+3cdot 2^{s-1} r+1$ is neither divisible by $latex r$, nor by $latex 2$ (the $latex +1$ appearing at the end is the culprit). The same argument works for any $latex sge 2$, since then $latex n$ must be divisible by $latex 4$ (since $latex x=4cdot 2^{s-2} r$ divides $latex n$) and we can use the fact that $latex

And this completes the proof of the special case of Fermat’s Lost Theorem.

*Pheeeeeewwww!*

But, what about the original question? What if $latex y$ is not equal to 1? Can anyone solve the general problem and claim eternal fame? Is the above proof correct, or is a dark abyss waiting for me? I look forward to your comments!

» Ένα «εύκολο» μαθηματικό πρόβλημα physics4meOctober 28, 2012 at 3:00 pm[…] έχει μια και μοναδική λύση για x>1 και m,n>1 Η λύση και το ιστορικό του προβλήματος βρίσκονται ΕΔΩ […]

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Giorgos TsichlisAugust 31, 2013 at 2:09 amYou say

“On the other hand, (2k+1)^m+1 always has remainder 2 when divided by 4”

This is not true when m and k are odd

As for Fermat’s last theorem,check this:

http://www.mymathforum.com/viewtopic.php?f=40&t=40857

and specially the August 3th post in this thread, that has the latest version

spirosAugust 31, 2013 at 3:37 amGiorgo, you are absolutely right! I went back to read why I wrote that sentence and realized that I went overboard when I asked (why?) without giving a little more of the proof beforehand. In my head, I had already concluded that m has to be even, since if m and x are odd, there is a contradiction: (x+1)^n = x^m+1 implies (x+1)^{n-1} = x^{m-1}-x^{m-2}+…-x+1. The l.h.s. is even, whereas the right hand side is odd (as a sum of an odd number of odd terms). I will update the post to reflect your sharp observation. But most importantly, thank you for reading the post so closely!!! And here I thought nobody cared… 🙂

giorgos tsichlisSeptember 13, 2013 at 11:44 pmCould I have your e-mail please? I may have something you might be interested in, but i don’t want it publicly known

The million dollar conjecture you’ve never heard of… | Quantum FrontiersSeptember 26, 2013 at 12:22 pm[…] a blog like this one and writing about imaginary stuff like Fermat’s Lost Theorem means that you get the occasional comment of the form: I have a really short proof of a famous open […]

Willie JohnsonNovember 14, 2013 at 5:35 pmThe solutions are pretty simple once you realize you are dealing with the Pythagorean theorem. To solve Beal you first solve Fermat’s Last theorem. Say A^n +B^n=C^n. Say n=4. Now what every body tries to do is take A,B,and C to the Fourth power. NO!!! In reality you have a^2A^2 +b^2B^2=c^2C^ 2 where a^2 ,b^2, and c^2 c are now dimensionless factors applied to the AREAS of A^2, B^2, and C^2. This is the basis for solving the Fermat equation. Also A and B can be expressed in terms of Ccos and Csin. So for the Beal you just take your x and y exponents and express them in terms of z and solve like you would the Fermat. I’ve got about 4 short margin proofs for the Beal. My book How To Solve the Beal and Other Mathematical Conjectures is free on Kindle Books for the next 3 days. The solutions really are simple.