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?
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$\begingroup$ If $a\mid b$, then $3^a-2^a$ is a divisor of $3^b-2^b$. $\endgroup$– Jack D'AurizioJan 8, 2017 at 18:26
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1 Answer
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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.
answered Jan 8, 2017 at 18:27
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2$\begingroup$ welp, I'm stupid. Thank you! I will remove this post in a bit to stop embarassing myself. Have a good day! $\endgroup$ Jan 8, 2017 at 18:44
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5$\begingroup$ 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. $\endgroup$ Jan 8, 2017 at 18:45
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1$\begingroup$ alright, a shame that this is my first post on this beautiful site. Amazing comunity as well! $\endgroup$ Jan 8, 2017 at 19:10