How to interpret result of manipulation of $Z^n-X^n$?

Question

Start from eq. (1) $Z^n-X^n=Y^n$, where $n, X, Y$ and $Z$ and are positive integers $X<Y<Z$. Replace $Z = X + B$, and remember eq. (2) $$X=Z-B$$ $B$ is also a positive integer. Then $$(X+B)^2Z^{n-2}-X^2X^{n-2}=Y^n.$$ Solve $X^2$: $$X^2=\frac{Y^n}{Z^{n-2}-X^{n-2}} - \frac{B(X+Z)Z^{n-2}}{Z^{n-2}-X^{n-2}}.$$ Divide by $X$: $$X = \frac{Y^n}{(\frac{X}{Z})Z^{n-1}-X^{n-1}} - \frac{BZ^{n-1}\frac{Z+X}{Z}}{(\frac{X}{Z})Z^{n-1}-X^{n-1}}.$$ Replace $\frac{X}{Z} = r$, $0<r<1$, and the result is eq. (3):$$X=\frac{Z(1-r^n)-B(1+r)}{r-r^{n-1}}$$ Now, this result does not seem to fit with the original assumption $X=Z-B$. Namely, according to the result $X$ is $Z-B$ only if the ratio $r$ is zero (which it cannot) in numerator but that would also cause the denominator to be zero. Please let me know how to interpret this (I am not a mathematician so I may sometimes be easily confusing/confused).

Answer 1

Remembering that $B = Z - X = Z(1 - r),$ a simpler way to derive your final equation is

\begin{align} X = rZ &= \frac{r^2 - r^n}{r - r^{n-1}} Z \\ &= \frac{(1 - r^n) - (1 - r^2)}{r - r^{n-1}} Z \\ &= \frac{Z(1 - r^n) - Z(1 - r)(1 + r)}{r - r^{n-1}} \\ &= \frac{Z(1 - r^n) - B(1 + r)}{r - r^{n-1}}. \end{align}

So the final equation indeed is something of the form $\frac{uZ - vB}{w}$, but you get it by slicing and dicing your terms (or whatever you want to call it; in any case it's a lot of algebraic manipulation). The only role of $X = Z - B$ is in the step $Z(1 - r) \to B.$

Similarly, we can confirm that $$ 3 = \frac{8\cdot 11 - 5\cdot 14}{6}, $$ where there is no sensible way of canceling the $6$ with either the $11$ or the $14$ in order to derive the equation $3 = 8 - 5$, but nevertheless it is true that $3 = 8 - 5$.

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As a side note, if $n = 2$ then $r^{n-1} = r,$ so anything with $r - r^{n-1}$ in the denominator is undefined. Really this should be no surprise, since then $n - 2 = 0$, implying that $X^{n-2} = Z^{n-2} = 1$ and $Z^{n-2} - X^{n-2} = 0,$ so dividing by $Z^{n-2} - X^{n-2}$ is not legitimate.

Since it has already been observed that $n \leq 2$ by Fermat's Last Theorem, it follows that the only positive integer value of $n$ for which $Z^{n-2} - X^{n-2} \neq 0$ is $n = 1.$ In that case your final equation comes out to

$$ X = \frac{Z(1 - r) - B(1 + r)}{r - 1} = \frac{B(1 + r) - Z(1 - r)}{1 - r} = B\frac{1 + r}{1 - r} - Z, $$

which looks even worse at first glance ("wrong signs" on the $B$ and $Z$ terms), but note that

$$ B\frac{1 + r}{1 - r} = B + \frac{2rB}{1 - r} = B + \frac{2rZ(1-r)}{1 - r} = B + 2X. $$

And it makes perfect sense that $X = 2X - (Z - B).$

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So what's really going on here?

It is a fact that if you have any two integers $P$ and $Q$, you can choose any integer multiple of $\gcd(P,Q)$ that you like, and there will be a linear combination $uP + vQ$, where $u$ and $v$ are integers, that comes out to that multiple of $\gcd(P,Q)$. Moreover, there are infinitely many such linear combinations, because if $$ uP + vQ = k \gcd(P,Q) $$ then $$ u'P + v'Q = k \gcd(P,Q) $$ where $u' = u + Q/\gcd(P,Q)$ and $v' = v - P/\gcd(P,Q)$. This also works if $u' = u - Q/\gcd(P,Q)$ and $v' = v + P/\gcd(P,Q)$, and it's a simple consequence of the fact that $QP - PQ = 0.$

This works as well for rational multiples of $\gcd(P,Q)$ where $u$ and $v$ are rational.

Applying this to your equations, let $P = Z$ and let $Q = B.$ Then $X = Z - B$ is a rational multiple of $\gcd(P,Q)$. So is $X(r - r^{n-1})$. Obviously we can make this multiple of $\gcd(P,Q)$ by scaling the linear combination $Z - B$: $$ X(r - r^{n-1}) = Z(r - r^{n-1}) - B(r - r^{n-1}). $$ That is, we have the linear combination $uP + vQ$ where $u = r - r^{n-1}$ and $v = r^{n-1} - r.$ But that is only one of the infinitely many possible linear combinations of $Z$ and $B$ with the same total. By increasing $u$ by a suitable amount and reducing $v$ by a suitable amount we can obtain $$ X(r - r^{n-1}) = Z(1 - r^n) - B(1 + r). $$ In the case where $r \neq r^{n-1}$ you can divide through by $r - r^{n-1}$ and obtain your Equation $(3)$.

What does this have to do with the equation $Y^n = X^n + Z^n$? Not much. It is all simply a consequence of the assumption that $X$ and $Z$ are integers. Notice that the only appearance of $Y$ in any of your equations is in the form $Y^n,$ and except for the initial assertion that $Y$ is an integer you never again rely on the fact that $Y$ is an integer; after a while you replace $Y^n$ by $Z^n - X^n$ and then it's gone. So really all of these algebraic manipulations are just about the integers $Z,$ $X,$ and $B,$ and as we see here there is nothing at all remarkable about the fact that we're able to recombine $Z$ and $B$ in a linear combination that is not just $Z - B$ and get the total $X$ again. It's just a simple consequence of the integer equation $X = Z - B.$

In particular, none of this tells us anything about the solutions to $Y^n = Z^n - X^n.$ It does not in any way contradict the hypothesis that an integer solution to this equation exists when $n = 3$ or when $n$ is any larger integer. And that's a good thing, because if there were a contradiction, what would it say about the case where $n = 1$?

Of course we still know what we know about the integer solutions of $Y^n = Z^n - X^n,$ thanks to some more advanced mathematics. All of the arithmetic gymnastics in the question and in this answer do not contradict Andrew Wiles's work; by the same token, they do not lead to a simpler demonstration of Fermat's Last Theorem either.