This was a question from a previous year in a test and I couldn't solve it yet.

If $3^n - 2^n$ is prime, then $n$ must be prime.

Do you have any tips, suggestions?

  • $\begingroup$ If $a\mid b$, then $3^a-2^a$ is a divisor of $3^b-2^b$. $\endgroup$ – Jack D'Aurizio Jan 8 '17 at 18:26

Suppose that $n$ is composite, say $n = ab$. Then $$ 3^n - 2^n = (3^a)^b - (2^a)^b,$$ and $$x^b - y^b = (x -y) (x^{b-1} + x^{b-2}y + \cdots + y^{b-1}).$$ Taking $x = 3^a$, $y = 2^a$ gives a non-trivial factorization of $3^n - 2^n$. Hence if $n$ is composite, then $3^n - 2^n$ is as well.

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  • 1
    $\begingroup$ welp, I'm stupid. Thank you! I will remove this post in a bit to stop embarassing myself. Have a good day! $\endgroup$ – J. Dionisio Jan 8 '17 at 18:44
  • 4
    $\begingroup$ No, don't remove the post. Your answerer spent effort and got 5 upvotes for it. Don't rob him of his points. Embarasmunt is good for us. $\endgroup$ – B. Goddard Jan 8 '17 at 18:45
  • $\begingroup$ And so is embarrassment. $\endgroup$ – Robert Israel Jan 8 '17 at 18:54
  • $\begingroup$ alright, a shame that this is my first post on this beautiful site. Amazing comunity as well! $\endgroup$ – J. Dionisio Jan 8 '17 at 19:10

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