<s>[INST] Finding positive integer(s) $n$ for which $3^{n-1} + 5^{n-1} | 3^{n} + 5^{n}$: Is my solution correct? The problem is the following

Find all positive integers $n$ such that $3^{n-1} + 5^{n-1}$ divides $3^{n} + 5^{n}$.

And I attempted as the following:
As  $3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n}$,
$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 5(3^{n-1} + 5^{n-1}) = -2\cdot 3 ^ {n-1}$$
And
$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 3(3^{n-1} + 5^{n-1}) = 2\cdot 5 ^ {n-1}$$
These expressions say that
$$2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  p \in \mathbb{Z}^{+}$$
And
$$2\cdot 5 ^ {n-1} = q(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  q \in \mathbb{Z}^{+}$$
By addition, it becomes clear that $q = 2 - p$.
Hence, the last expression is equivalent to $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$.
Solving for $p$, it is equal to $\frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$.
Here, $p$ is odd only for $n = 1$, hence it is the only solution.
Is this attempt a solution for the problem? And for the effort, Thanks!
 [/INST] What you did is mostly correct, but your argument that $p = \frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$ is only odd for $n = 1$ is not true since for all odd $n$, the result is either not an integer or it's an odd integer. Instead, $p$ is odd, as you stated, but with $2 - p \gt 0$ from $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$ since the left side is positive, the right side must be as well, means only $p = 1$ works. Thus, $2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1})$ gives $3^{n-1} = 5^{n-1} \implies n = 1$, so your solution is correct.
A somewhat easier way to see this is that, from what you wrote, $3^{n-1} + 5^{n-1} \mid -2\cdot 3^{n-1} \implies 3^{n-1} + 5^{n-1} \le \left|-2\cdot 3^{n-1}\right| = 2\cdot 3^{n-1}$. For $n \gt 1$, we have $5^{n-1} \gt 3^{n-1} \implies 3^{n-1} + 5^{n-1} \gt 2\cdot 3^{n-1}$, so only $n = 1$ is a possible solution. Since $n = 1$ gives $2 \mid 8$, this means only $n = 1$ works.
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