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Assume $n > 1$ and $n$ is odd because it's easy if $n$ is even.

Please help prove this.

$x^n + y^n = c$ has finitely many integral solutions if $c \neq 0$?

Thank you all for replying. I think I've just found my answer to this one. And I think it's quite simple. Here it goes.

(Non-negative solutions mean solutions $(x,y)$ where $x, y \geq 0;$ and negative solutions mean solutions $(x, y)$ where $x, y < 0$.)

The equation has finitely many non-negative solutions as well as negative solutions. This is quite trivial to show. From here, consider only the solutions $(x, y)$ where $x$ and $y$ have opposite signs. Without loss of generality, assume $x >0$ and $y <0$.

Let $z = -y >0$. So the equation becomes $x^n - z^n = c$.
$\implies |x^n - z^n| = |x-z||x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}|=|c|$.

Since $|x-z| \geq 1$ (if $|x-z| = 0, c = 0$), $|c| \geq |x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}| \geq |x^{n-1} + z^{n-1}| = x^{n-1} +z^{n-1}$

Now, it's obvious that there are only finitely many such $(x, z)$. And it's pretty much done here. Please let me know of my any mistake.

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  • $\begingroup$ Are we given anything about $c$ ? $\endgroup$
    – Henry
    Dec 3, 2014 at 8:32
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    $\begingroup$ Perhaps this helps: $(x+1)^n-(x^n) > nx^{n-1}$ $\endgroup$ Dec 3, 2014 at 11:08

3 Answers 3

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Let's look at the factorization of $x^n+y^n$.

For all odd $n$, $$x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+\cdots-xy^{n-2}+y^{n-1}\right)$$

Thus, since $x^n+y^n=c$, $$\left(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1} \right)\Bigg| c$$

This implies that $$\left|x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\right|\leq |c|$$

Now here's the key to the argument: we claim that $$x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\geq \min\left(x^{n-1}, y^{n-1} \right)$$

If we can show this, then we will have shown that in order for $x^n+y^n=c$, we must have that $$\min\left(x^{n-1},y^{n-1}\right)\leq |c|$$

This wins the proof for us because this, combined with our restriction that $x^n+y^n=c$, is only true for finitely many $x$ and $y$. So all we have to do now is show the inequality holds. I'll do the $n=5$ case so you can see the general principal at work.

So we're considering the equation $$x^4-x^3y+x^2y^2-xy^3+y^4$$

If $|x|=|y|$ (this implies that $x=y$ since $x\neq -y$ because $x^n+(-x)^n=0$ and $c\neq 0$), then it's very easy to show that the inequality holds. $\checkmark$

Without loss of generality, let's let $|x|>|y|$. We're gonna get a few cases.

Case 1 $x\geq 0$ and $y\geq 0$. Then we have that $x^4-x^3y\geq 0$ and $x^2y^2-xy^3\geq 0$, both since $x>y$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4= \min\left(x^4,y^4\right) \checkmark$$

Case 2 $x\leq 0$ and $y\leq 0$. This case is extremely similar to Case 1. $\checkmark$

Case 3 $x\geq 0$ and $y\leq 0$. Then we have that $-x^3y\geq 0$ and $-xy^3\geq 0$, both since $y\leq 0$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4=\min\left(x^4,y^4\right) \checkmark$$

Case 4 $x\leq 0$ and $y\geq 0$. Super similar to Case 3. $\checkmark$

These are all of the possible cases, so that completes the proof for $n=5$. (Woohoo!)

Showing that that inequality holds for all odd $n$ is done using the same general idea as in the $n=5$ case. Then, once you've shown that, there are only finitely many $x$ and $y$ that can satisfy $$\min\left( x^{n-1},y^{n-1}\right)\leq |c|$$ and $$x^n+y^n=c$$ so we're finished!

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  • $\begingroup$ This shows that all of the solutions for $x$ and $y$, for all $n\ge2$ put together, are still a finite number. $\endgroup$
    – Empy2
    Dec 3, 2014 at 11:41
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For a given $c \neq 0$ and $n >1$, the set $\mathcal{S}_{n,c} = \{ (x,y) \in \mathbb{Z}^2, x^n+ y^n = c\}$ can be represented as $$ \mathcal{S}_{n,c} = \mathcal{S}_{1}\cup \mathcal{S}_{2} \cup\mathcal{S}_{3}\cup \mathcal{S}_{4},$$

where

$\mathcal{S}_{1} = \{ (x,y) \in \mathbb{N}^2, x^n+ y^n = c\}$;

$\mathcal{S}_{2} = \{ (x,y) \in \mathbb{N}^2, x^n- y^n = c\}$;

$\mathcal{S}_{3} = \{ (x,y) \in \mathbb{N}^2, x^n- y^n = -c\}$;

$\mathcal{S}_{4} = \{ (x,y) \in \mathbb{N}^2, x^n+ y^n = -c\}$.

Notice that if $c > 0$, $\mathcal{S}_{4} = \emptyset$, and if $c < 0$, $\mathcal{S}_{1}= \emptyset$. Thus, for a given $c$, we can eliminate a set by assigning a sign to $c$. The sets $\mathcal{S}_{2}$ and $\mathcal{S}_{3}$ are interchangeable with the sign of $c$. So, we can choose $c >0$. Thus, $$ \mathcal{S}_{n,c} = \mathcal{S}_{1}\cup \mathcal{S}_{2} \cup\mathcal{S}_{3}.$$

Because the set $\mathcal{V}_{n,c} = \{ (x,y) \in \mathbb{N}^2, |x^n-y^n| \leq |c|\}$ is finite, and $\mathcal{S}_{i}\subseteq \mathcal{V}_{n,c}$, for all $i \in \{1,2,3\}$, then $\mathcal{S}_{n,c}$ is also finite.

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Without loss of generality we can assume $c > 0$. Let's look what will happen if $c < |x| < |y|$. Then (obviously $y > 0$): $$ |y^n + x^n| \geq |y^n| - |x^n| \geq |y|^n - (|y| - 1)^n \geq (|y|^{n-1} + |y|^{n-2}(|y| - 1)+ \dots) > c $$

Last case if $|x| = |y|$. Then if they are equal positive numbers then $x^n = \frac{c}{2}$ has only $1$ solution and otherwise since $c\neq0$ we do not have solutions.

Finally we got that we can have solutions iff $|x| \leq |y| \leq c$ but there are finitely many such $(x, y)$.

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