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?
Proof that if $3^n - 2^n$ is prime then $n$ is prime
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elementary-number-theory
prime-numbers
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0If $a\mid b$, then $3^a-2^a$ is a divisor of $3^b-2^b$. – 2017-01-08
1 Answers
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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.
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1welp, I'm stupid. Thank you! I will remove this post in a bit to stop embarassing myself. Have a good day! – 2017-01-08
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4No, 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. – 2017-01-08
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0And so is embarrassment. – 2017-01-08
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0alright, a shame that this is my first post on this beautiful site. Amazing comunity as well! – 2017-01-08