Determine all positive integers $n$ such that:

$$
5^{n-1}+3^{n-1} \mid 5^{n}+3^{n}
$$
Let $n \in \mathbb{N}$. We have:

$$
5^{n-1}+3^{n-1} \mid\left(5^{n}+3^{n}\right)-5\left(5^{n-1}+3^{n-1}\right)
$$

Then

$$
5^{n-1}+3^{n-1} \mid 3^{n}-5 \times 3^{n-1}
$$

i.e.

$$
5^{n-1}+3^{n-1} \mid 3^{n-1}(3-5)
$$

. We also have $\operatorname{gcd}\left(5^{n-1}+3^{n-1}, 3^{n}-1\right)=1$: let $d$ be a common divisor of $5^{n-1}+3^{n-1}$ and $3^{n-1}$, then $d \mid 3^{n-1}$ and $d \mid 5^{n-1}$, so $d$ is both a power of 3 and of 5, i.e., $d=1$.

Next, by Gauss's lemma,

$$
5^{n-1}+3^{n-1} \mid 2
$$

and thus $5^{n-1}+3^{n-1}=2$ and $n=1$.

Conversely, $n=1$ works.