For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$?

Question

Stuck on this problem for some time: For which $n ≥ 0$ is $2^n + 2 · 3^n$ divisible by $8$? I've reached the conclusion that $n = 1$ is the only solution to the question at hand, but I cant quite prove that/why this is the case.

The properties of the evaluation of the expression as $n = 1$, $n = n + 1$ gives that the first term increases by a multiple of $2$, while the second by a multiple of 3, how can this help proving the solution?

Showing that $2^n + 2 · 3^n \equiv 0$ (mod 8) only when $n = 1$, why is it not enough to show the cases $0...7$? How would I go about proving the solution by modular arithmetic?

Going by induction, what would be the best way to prove that indeed 1 is the only solution - that the expression is not divisible by $8$ for any $n \geq 1$, in that case - how would this look? Or is there a better, different angle to attack it?

Answer 1

To elaborate on the argument presented in the comments. Let $a_n=2^n+2^*3^n$. We are only interested in $a_n\mod (8)$.

We note that $n≥3\implies 2^n\equiv 0\mod (8)$. So, for $n≥3$ we have $a_n\equiv 2^*3^n\mod (8)$. But it is clear, from unique factorization, that $8$ never divides $2^*3^n$ (indeed it is easy to see that $2^*3^{2n}\equiv 2$ and $2^*3^{2n+1}\equiv 6$). All that remains is to check $a_0, a_1,a_2$ to see that only $a_1$ is divisible by $8$.

Answer 2

If $n$ is even, then $3^n \equiv 1 \bmod 8$ and so $2^n + 2 \cdot 3^n \equiv 2^n + 2 \not\equiv 0 \bmod 8$ because it is twice an odd number.

If $n$ is odd, then $3^n \equiv 3 \bmod 8$ and so $2^n + 2 \cdot 3^n \equiv 2^n + 6 \not\equiv 0 \bmod 8$ because it is twice an odd number if $n>1$.

Answer 3

If $n\geq 3$, $2^n+2\cdot 3^n$ is either $2$ or $6\pmod{8}$, since $3^2=9\equiv 1\pmod{8}$ and $8\mid 2^n$.

So we just have to check by hand the cases $n\in\{0,1,2\}$.