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Let $\mathbb{N}$ denote the set of positive integers.

We define a relation $R\subseteq \mathbb{N}^3$ by

$$ R = \{(x,y,z) \in \mathbb{N}^3: \exists n\in \mathbb{N}: 1< n \leq \max\{x,y,z\} \land \exists k\in \mathbb{N}: x^k + y^k \equiv z^k (\text{mod } n)\}.$$

What is an example of an element of $\mathbb{N}^3\setminus R$?

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    $\begingroup$ For any $(x,y,z)$ (except those of the form $(x,1,x)$ or $(1,y,y)$) can't you just choose $k$ large and $n=\vert x^k+y^k-z^k\vert$? $\endgroup$
    – Jeremy Rickard
    Commented Jun 29, 2015 at 7:19
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    $\begingroup$ Well, certainly for $n=\vert x^k+y^k-z^k\vert$, $x^k+y^k\equiv z^k\pmod{n}$, and for fixed $(x,y,z)$, with the exceptions noted, $n\to\infty$ as $k\to\infty$, so $n>\text{max}\{x,y,z\}$ for large $k$. I wondered if you had the inequality backwards, or something? $\endgroup$
    – Jeremy Rickard
    Commented Jun 29, 2015 at 7:33
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    $\begingroup$ What is the point of the inequality? $a + m n \equiv a \pmod{n}$. So $x,y,z$ can be arbitrary large and the congruence is satisfied if a single solution exists. $\endgroup$
    – joro
    Commented Jun 29, 2015 at 8:31
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    $\begingroup$ For $(x,y,z)\in\mathbb{N}^3$. If there is a prime $n$ dividing all or just one of the three then $(x,y,z)\in R$. (Take $k=n-1$ for the second case.) $\endgroup$
    – Chris Wuthrich
    Commented Jun 29, 2015 at 12:11
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    $\begingroup$ And an answer to your question is $(2,2,1)$. In summary, this question does not sound like a research level question. $\endgroup$
    – Chris Wuthrich
    Commented Jun 29, 2015 at 13:04

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