$n \in \mathbb N$, find all values of $n$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.

Question

Find all values of $n \in \mathbb N$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.

Let $k=3^{2n+1}-2^{2n+1}-6^n$. By testing with numbers I got that $k$ cannot be prime for any value of $n$ (tested to $n=20$, so I assumed it's same for every value of $n$).

For $n=1$, $k<0$ $\implies$ ~~$k$ is not prime.~~ (EDIT: oh... it's actually prime if $n=1$. Stupid me calculated $3^3=9$)

For $n>1$:
I managed to prove that if $n$ is even, $k$ is not a prime:
If $n$ is even, then $$3^{2n+1}\equiv3\ (mod\ 10),\quad 2^{2n+1}\equiv2\ (mod\ 10),\quad 6^n\equiv6\ (mod\ 10)$$ $$\implies k\equiv5\ (mod\ 10)$$ So $k$ is divisible by 5 and therefore not prime.

Using the same method for if $n$ is odd, I get: $k\equiv3\ (mod\ 10)$, which doesn't prove anything.

How do I solve for when $n$ is odd?

Have you tried factoring. Often these "tricks" involve that the expression can be factored and hence can't be prime unless one the terms is $1$. — fleablood, Dec 19 '18 at 20:22
See my answer in the dupe where I explain how to derive the answer (vs. pull it out of a hat like magic). — Bill Dubuque, Dec 19 '18 at 20:41

nonuser · Accepted Answer · 2018-12-19T20:42:30.377

4

Let $x= 3^n$ and $y=2^n$, then we have $$3x^2-xy-2y^2 = (3x+2y)(x-y)$$

so this expresion is prime if $x-y=1$ (since $3x+2y>1$) that is if $3^n = 2^n+1$, so $$ 3^n-2^n = 1\implies (3-2)(3^{n-1}+3^{n-2}\cdot 2+....+2^{n-1})=1$$

But this holds only if $n=1$.

edited Dec 19 '18 at 20:42

answered Dec 19 '18 at 20:22

nonuser

90,026

1

It's prime "only if" $x - y=1$ (if we assume $x > y$ and $3x + 2y > 1$... which are safe assumptions). But it's not "if and only if" as $3x +2y$ needn't be prime for all we know yet. – fleablood Dec 19 '18 at 20:30
I explain how to derive this from scratch in this old answer. – Bill Dubuque Dec 19 '18 at 20:54

$n \in \mathbb N$, find all values of $n$ for which $3^{2n+1}-2^{2n+1}-6^n$ is a prime number.

1 Answers1