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Find all positive integers n, such that \begin{equation} 22^n+8^n-11^n-3^n \equiv 0\ (\textrm{mod}\ 209) \end{equation} My attempt so far: \begin{equation} 11^n(2^n-1)\equiv 0\ (\textrm{mod}\ 209) \Leftrightarrow\ (2^n-1)\equiv 0\ (\textrm{mod}\ 19) \end{equation} But from Euler theorem \begin{equation} \Rightarrow\ 2^{18}\equiv 1\ (\textrm{mod}\ 19) \Rightarrow\ 2^{180}\equiv 1\ (\textrm{mod}\ 19). \end{equation} In the same fashion we get that \begin{equation} 8^{180}-3^{180}\equiv 0\ (\textrm{mod}\ 209). \end{equation} Finally, \begin{equation} \Rightarrow\ 11^{180}(2^{180}-1)+8^{180}-3^{180}\equiv 0\ (\textrm{mod}\ 209). \end{equation} But maybe there are even more solutions, so how can I find them?

Bill Dubuque
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Jut3
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  • Try to think of a more concise title to your questions. You are trying to determine the cardinality of a subset of the natural numbers, is it finite or infinite? might be a good preliminary question that leads to the final answer you are after here. – Adam Ledger Mar 08 '21 at 20:04
  • @AdamL If you say so – Jut3 Mar 08 '21 at 20:05
  • SPecial case of the general methoids explained in the linked dupes (and their links). – Bill Dubuque Mar 09 '21 at 09:49

1 Answers1

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I'm not sure why you are partitioning the left side, as you are eliminating solutions. All we want is the expression to be equivalent to $0$ mod $11$ and mod $19$.

For mod $11$, we have $$8^n\equiv3^n\bmod{11}\to(-3)^n\equiv3^n\bmod{11}$$So the solutions to this mod equation are when $n$ is even.

For mod $19$, we have $$3^n+8^n-11^n-3^n\equiv0\bmod{19}\to8^n\equiv(-8)^n\bmod{19}$$Which again is true for all even $n$ (and only for even $n$).

So, the system is true for any even $n$.

Rushabh Mehta
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