Find all positive integers $n$ such that $n$ divides $9^{n-1}+3^{n-1}+1$.
I suspect there are no solutions other than $n=1$. Assuming $n$ divides it gives $3^{3(n-1)} \equiv_n 1$, from which one might start arguing using orders mod $n$? I'm unsure of how to proceed with this observation.
Comment: We rewrite the expression as:
$A_n=9^{n-1}+3^{n-1}+1=(3^{n-1}+1)^2-3^{n-1}$
If n is odd, $n=2k+1$ then we have:
$A_n=(3^{2k}+1)^2-3^{2k}=(3^{2k}+3^k+1)(3^{2k}-3^k+1)$
That is for odd n the expression is definitely a composite.
We can use this for reducing the expression to a multiple of expression of lower power; we can write:
$A_{2k+1}=B\times A_k$
where $B=(9^k-3^k+1)$
For example:
$A_{35}=B\cdot (9^{17}+3^{17}+1)$
$A_{17}=B\cdot (9^8+3^8+1)$
$(9^{8}+3^8+1)=A_9$
$A_9=B\cdot(9^{5}+3^5+1)$
$A_9=6643\times 6481=7\times 13\times 73\times 6481$
That is $A_{35}$ is a multiple of $A_9$. Also we can see that for $n=35$ there is a common divisor $7$ in $n$ and $A_n$. We can find many such $n$ and claim that there are infinite n such that there are some common divisors in $n $ and $A_n$.That is ,at most, $n$ and $A_n$ may have some common divisors. We may also ask: Using this property, can we construct a number like n such that ,$n|9^n+3^n+1$, ?