Having difficulty in proving this: $1989\mid n^{n^{n^{n}}} - n^{n^{n}}$ for all $n \in \Bbb N$.

Prime factorization of $1989$ is $3^2 \times 13 \times 17$.

Please Help!

  • 5
    $\begingroup$ TeX tip: Use \mid for "divides" instead of |. Compare $a \mid b$ to $a|b$. And of course there's also \nmid: $a \nmid b$. $\endgroup$
    – kahen
    Commented Mar 2, 2015 at 11:27
  • 1
    $\begingroup$ @user8795: I think you need to modify your conjecture to $1989 \mid n^{n^{n^n}}-n^{n^n}$ for all integer $\color{blue}{n>2}$. I've tested it with Wolfie and it is true for $2<n<20$ so far. If it is false, then it should be interesting to know the first $n>2$ that it fails. $\endgroup$ Commented Jul 31, 2015 at 8:53
  • $\begingroup$ I have posted an answer to this related (almost duplicate) question. I do not know whether it is correct, though. $\endgroup$
    – Τίμων
    Commented May 29, 2016 at 21:43

1 Answer 1


This wrong for $n=2,\;$ because $1989$ does not divide $2^{2^{2^{2}}} - 2^{2^{2}}=2^{16}-16=65520$.


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