Let $x, y, p, n, k$ be positive integers such that $n$ is
odd and $p$ is an odd prime. Prove that if $x^n + y^n = p^k$
, then $n$ is a power of $p$.
By the first law of LTE,we get
$V_p(x^n-y^n) = V_p(x-y)+V_p(n)=k$
From there, we have
$ V_p(n)=k- V_p(x-y)$
Therefore, it would be enough if i could show that
$n=p^{(k-V_p(x-y))}$
I am stuck here and would appreciate hints or if i made any mistakes here i would like them to be pointed out.
Number theory is not my area of specialty, and I didn't know about the LTE (Lifting the Exponent) Lemma, but I've now done some research on it. It's discussed in MSE's What can I do with the lifting the exponent lemma?, as well as Lifting The Exponent Lemma, where it's also proven & it links to Lifting The Exponent Lemma for Sums which is what you need for adding $2$ powers instead. As far as I can tell, apart from using $-y$ instead of $+y$, what you've done is fine. However, note that the LTE is, as far as I can tell, just an inclusion technique, i.e., once you know what prime factors you have, then you can determine their powers, but it's doesn't provide any specific exclusion. For your particular question, you basically need to prove when the only prime factor of the sum is $p$, then the only prime factor of the power is also $p$. As such, I don't see how you can use LTE to prove your question but, if I'm wrong, I hope somebody else will answer here to show us how it can be done.
Instead, the following is a proof using just fairly basic number theory and the binomial theorem. The provided equation is
$$x^n + y^n = p^k \tag{1}\label{eq1}$$
First, note that if $p \mid x$ or $p \mid y$, then $p$ must divide both $x$ and $y$. You can then reduce \eqref{eq1} by dividing both sides by $p^n$ to then use $x_1 = \frac{x}{p}$ and $y_1 = \frac{y}{p}$, with $k_1 = k - n$, in an equation of the same form as \eqref{eq1}. This process can be continued until we get to $p \nmid x$ and $p \nmid y$ (or you could have more directly just determined that each power of $p$ in $x$ and $y$ must be the same), so I will next check this base case. First, since $n$ is odd, note that $x + y \mid x^n + y^n$ and, thus, $x + y \mid p^k$. As $x$ and $y$ are positive integers, this means that
$$x + y = p^j \text{ for some } j \in \mathbb{N} \tag{2}\label{eq2}$$
Next, assume that $n$ is not an integral power of $p$. Thus, there must be an odd prime, other than $p$, say $m$ such that $m \mid n$. Note also that $x^m + y^m \mid x^n + y^n$, so $x^m + y^m \mid p^k$ and, thus,
$$x^m + y^m = p^i \text{ for some } i \in \mathbb{N} \tag{3}\label{eq3}$$
From \eqref{eq2}, substitute $x = p^j - y$ into \eqref{eq3} and expand using the binomial theorem to get
$$p^{jm} - mp^{j\left(m - 1\right)}y + \ldots - \frac{m\left(m - 1\right)}{2}p^{2j}y^{m-2} + mp^j y^{m-1} - y^m + y^m = p^i \tag{4}\label{eq4}$$
Now, divide both sides by $p^j$ to get
$$p^{j\left(m - 1\right)} - mp^{j\left(m - 2\right)}y + \ldots - \frac{m\left(m - 1\right)}{2}p^{j}y^{m-2} + my^{m-1} = p^{i-j} \tag{5}\label{eq5}$$
As $m \gt 2$, with $x$ and $y$ being positive integers, with not both being $1$, you can see from \eqref{eq2} and \eqref{eq3} that $i \gt j$. Thus, the RHS of \eqref{eq5} is a positive power of $p$. However, all the terms on the LHS of \eqref{eq5} have an explicit factor of $p$ except for the last one, i.e., $my^{m-1}$, so $p \mid my^{m-1}$. However, it's given that $p \nmid m$ and $p \nmid y$, so this is not possible. Thus, the original assumption of an $m$ existing must be false. As such, the only prime factors of $n$ is just $p$, giving that $n$ is a power of $p$.