# Fermat’s Lost Theorem

Pierre Fermat, known for his Last Theorem and for rarely proving any of his claims.

Last week, I posted a series of increasingly difficult challenges in the post One line proof. Today, I would like to spend some time with the second challenge:

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

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 $y=1$ and $x > 1$. I expect that many of you know the following variant of the problem:

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

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.

And so Pythagoras called it “hypotenuse”, cause he thought that math should be fun.

Now, if you look at the statement of the theorem, you will notice that $m$, the power to which the integers are raised, has to be greater than 2 for the theorem to work. Why? Well, you may have seen this equation before: $z^2=x^2+y^2$. It was discovered by my great-great-great-great-…-great uncle, Pythagoras of Samos, while studying the length of the longest side (hypotenuse) of right triangles. These days it goes by the name of Pythagorean Theorem. In fact, the equation $z^2=x^2+y^2$ has an infinite number of solutions in integers given by: $x = m^2-n^2, y = 2mn, z=m^2+n^2$, where $m,n$ are positive integers (verify that these solutions satisfy the Pythagorean equation – you only need to know that $(a+b)^2 = a^2+2ab+b^2$). And those are all of the solutions to the equation.

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 $z=(x+y)^p$ and $m=n\cdot p$. 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 $(x+1)^n=x^m+1^m$ has a unique solution for $x > 1$ and $m,n > 1$.

Solution: We will need the following important equation which holds for any number $x$ and integer $n$: $x^n-1 = (x-1)(x^{n-1}+x^{n-2}+\ldots+x^2+x+1)$. The proof of this identity is simple: Multiply each term in the second parenthesis with $x$ to get $x^n+x^{n-1}+x^{n-2}+\ldots+x^2+x$ and subtract $x^{n-1}+x^{n-2}+\ldots+x^2+x+1$ to get $x^n-1$ (since every other term cancels in the telescoping sum!) We have shown that the fraction $\frac{x^n-1}{x-1}$ is always equal to $x^{n-1}+x^{n-2}+\ldots+x^2+x+1$. 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

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

this implies that:

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

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

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

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

We are now ready to finish the job. Let $x=2k+1$ be an odd integer. Then, $(x+1)^m = 2^m(k+1)^m$ is divisible by 4 (since $m\ge 2$). On the other hand, $(2k+1)^m+1$ always has remainder 2 when divided by 4 (why?) So, $x$ cannot be an odd number and we are left with $x=2^s\cdot r$, for $s\ge 1$ and $r$ an odd integer. This makes our equation look like this:
$(2^s\cdot r+1)^{n} -1 = 2^{sm}\cdot r^{m}$. Let us look at the two smallest values of $n$, recalling that $x$ divides $n$: The smallest value is $n=2$, which implies $x=2$ and $m=3$. The next value is $n=4$, which gives: $(2^sr+1)^4-1 = 2^{4s}r^4+4\cdot2^{3s}r^3+6\cdot 2^{2s}r^2+4\cdot 2^sr$. The crucial step comes next: Factor out $2^{s+2}r$ from the expression on the right to get: $(2^s\cdot r+1)^{4} -1 = (2^{s+2} r) (2^{3s-2}r^3+2^{2s} r^2+3\cdot 2^{s-1} r+1) = 2^{sm}\cdot r^m$. Now what? Well, we are done! For the last two expressions to be equal, we must have that $r = 1$ and $s=1$, otherwise $2^{3s-2}r^3+2^{2s} r^2+3\cdot 2^{s-1} r+1$ is neither divisible by $r$, nor by $2$ (the $+1$ appearing at the end is the culprit). The same argument works for any $s\ge 2$, since then $n$ must be divisible by $4$ (since $x=4\cdot 2^{s-2} r$ divides $n$) and we can use the fact that $[(2^s+1)^4]^{n/4}-1^{n/4}$ has $(2^s+1)^4-1$ as one of its factors (recall the identity for $x^n-1$ near the beginning of the proof?), which is not divisible by large enough powers of $2$ and $r$. We can do the same trick, but now with $r\ge 3$ to show that $(2^s+1)^r-1$ is not divisible by $r^3$. Since $m \ge n \ge r \ge 3$, this implies that $r^3$ should divide $(2^s+1)^r-1$ and we have a contradiction.
And this completes the proof of the special case of Fermat’s Lost Theorem.

Pheeeeeewwww!

But, what about the original question? What if $y$ 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!

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