# Correct statement of Fermat's Last Theorem

I'm looking at the wikipedia page on Fermat's Last Theorem

In the statement it requires $a,b,c$ to be positive integers. Is that correct? I always took it to be no solutions in non-zero integers. But this wiki page makes a big deal out of the bases being positive. Has some counter-example turned up using negative integers that I'm missing? Otherwise, I think we should fix the wiki page.

• $a^n+0^n=a^n$ for any $a\in\mathbb{Z}$ and $n\geq 3$. – JP McCarthy May 13 '15 at 15:55
• @JpMcCarthy I think you misunderstood my question. Did you miss the word "non-zero"? – Gregory Grant May 13 '15 at 15:57
• If $n$ is even, then sign doesn't not matter. If $n$ is odd then put the negative number on the other side (eg $(-x)^n + y^n =(-z)^n \Leftrightarrow z^n + y^n =x^n$). – user10676 May 13 '15 at 15:59
• Because a solution in Z would yield one in N. – quid May 13 '15 at 16:02
• @GregoryGrant: Because the version stated is the version that most people will identity as "Fermat's Last Theorem". – Thomas May 13 '15 at 16:11

The formulations are equivalent. This is clear when $n$ is even (because then $x^n=(-x)^n$), so assume $n$ is odd. I will prove that if FLT has no positive solutions, it will have no non-zero solutions. We have few cases:

1. $a,b,c>0$ - we know this doesn't have solutions.

2. $a,b,c<0$ - if these were a solution, we would have $a^n+b^n=c^n$ and then by multiplying by $(-1)^n$ we would have $(-a)^n+(-b)^n=(-c)^n$ with $-a,-b,-c>0$.

3. $a,b<0, c>0$ - this is impossible, as then $a^n+b^n<0<c^n$

4. $a,b>0, c<0$ - same as above.

5. $a>0,b<0,c>0$ - if we have $a^n+b^n=c^n$, then $c^n+(-b)^n=a^n$.

Rest of the cases goes similarly, which I will leave for you.

• Got it thank you, I accept this explanation. – Gregory Grant May 13 '15 at 16:03
• Since the stronger statement (over $\mathbb Z$) is actually equivalent to the weaker (over $\mathbb N$), shouldn't the theorem always be stated over $\mathbb Z$? Therefore I still think the wikipedia page should be modified. Otherwise, why not just state it for prime exponents, since that's equivalent to the general case too? – Gregory Grant May 13 '15 at 16:08
• @GregoryGrant I think the reasons it's usually stated over $\Bbb N$ are purely historical. This is how Fermat (iirc) has stated it, and so we keep it. – Wojowu May 13 '15 at 17:44
• Thank you for your response. My Latin is poor, but my reading of Fermat's note is that he simply refers to there being no "numbers" that satisfy the equation. I don't think he specifies they must be positive. The accepted translation is "It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain." – Gregory Grant May 13 '15 at 17:54

We are considering the equation $$a^n + b^n = c^n.$$ Clearly if $n$ is even, then negatives just go away, so let's say that $n$ is odd.

You have some cases. One case is where $a$ and $c$ are positive, but $b$ is negative. Then the equation is equivalent to $a^n = (-b)^n + c^n$. So this case isn't interesting. You also have the case where $a$ and $b$ are positive and $c$ is negative. But there clearly is not such solution. If all are negative we also don't get anything new.

Anyway, if you continue considering the different cases you will realize that the important case is where $a$, $b$, and $c$ are all positive.

As @quid says above, if you find a solution in the non-zero integers, then you will have a solution in the natural numbers.

Suppose there is a solution in non-zero integers. If $n$ is even this immediately yields a soltution in positive integers.

So suppose $n$ is odd. If all three are negative, we clearly also get a positve solution. So assume one is negative two possitive (two negative, one positive can be reduced to this multipliying by $-1$). Clearly it cannot be $c$ that is negative, so assume it is $a$ but then we get $b^n =c^n -a^n = c^n + (-a)^n$ a positive solution.

In brief, insisting on positve is irrelevant, there are no nono-zero solutions either.

The person who had the best possible reason to think about the presentation of the conjecture is Andrew Wiles. In his paper on Fermat's Last Theorem, the formulation is symmetrical and allows rational solutions:

if $u^p + v^p + w^p = 0$ with $u,v,w$ rational and $p \geq 3$, then $uvw=0$.

It's interesting that he also chose to not use the traditional letters $x,y,z$ and $n$ for the variables in the problem. His paper "buries the lede" by not giving the elementary form of FLT until the sixth page, and stays as far away as possible from the archetypal Diophantine form of FLT other than to mention he has proved it, and to quote Fermat's formulation in Latin below the dedication.

• Thank you for this post. I also looked up Wiles paper and indeed he does state it over $\mathbb Q$. I have ten textbooks that state FLT. Nine of them state it over $\mathbb Z$. One of those actually claims Fermat originally stated it over $\mathbb N$ but that's not how I read Wiles quote. I have one book that states it over $\mathbb N$, Scharlau and Opolka's "From Fermat to Minkowski". Therefore I think the consensus in the math world is in contrast to the consensus on this page. – Gregory Grant May 13 '15 at 22:43
• @GregoryGrant, FLT over rationals and FLT over integers are equivalent by clearing denominators. In modern number theory, equations over rational numbers are often way more tractable than equations over integers, which is probably why Wiles state that it's over $\mathbb{Q}$. (After all, he proved modularity of elliptic curves over $\mathbb{Q}$, rather than $\mathbb{Z}$.) – Pig May 14 '15 at 0:50
• @user31814 Very true. In fact over $\mathbb Q$ you can go one step further and divide through both sides by $z^n$ to reduce it to a problem in two variables instead of three. In other words FLT $\Leftrightarrow$ $x^n+y^n=1$ has no non-trivial solutions in $\mathbb Q$ for $n\geq3$. – Gregory Grant May 14 '15 at 0:53