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: 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.
A: 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. 
A: 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.
A: 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:


*

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

*$a,b,c<0$ - 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 $-a,-b,-c>0$.

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

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

*$a>0,b<0,c>0$ - 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.
