$x^p+y^p$ is a primitive root modulo $p^2$ I can't solve the following problem:  

Let $p$ be a prime. Prove that there is a primitive root modulo $p^2$ of the form $x^p+y^p$.

Edit: There was a mistake, i hope now it's correct. Now the problem is very weak but i'm sure it's correct. My apologies if someone wasted his time with this problem.
 A: $\def\ZZ{\mathbb{Z}}$I've thought enough about this problem that I might as well write up an answer summarizing what I know. Let $p$ be an odd prime. Let $U$ be the group of units modulo $p^2$. As I assume you know, $U \cong \ZZ/p(p-1) \cong \ZZ/p \times (\ZZ/p)^{\times}$. Let's call these two factors $S$ and $T$. Explicitly, $S$ is the group of units of the form $1+kp$ and $T$ is the group of units of the form $x^p$.
For example, when $p=5$, the group $S$ is $\{ 1, 6, 11, 16, 21 \}$ and the group $T$ is $\{ 1, 32, 243, 1024 \} \equiv \{ 1, 7, 18, 24 \} \mod 25$. You can have fun checking that each of these really is a subgroup of $(\ZZ/25)^{\times}$.
An element of $U$ is a primitive root if it is not in $T$ and its projection to $T$ is a primitive root.
Let $X \subseteq U$ be the set of units of the form $x^p + y^p$. Notice that, if for $t=z^p \in T$, we have $t (x^p+y^p) = (zx)^p + (zy)^p$. So the set $X$ is invariant under translation by $T$. We see that $(s,t)$ is in $X$ if and only if $(s,t')$ is in $X$. Let $Y \subset S$ be the set of $s\in S$ such that $(s,t)$ is in $X$ for any, or equivalently every, $t \in T$. So $X = Y \times T \subset S \times T = U$.
An element $(s,t)$ in $X$ is a primitive root if and only if $s$ is not the identity of $S$ and $t$ is a primitive root. So the number of primitive roots you are interested in is $| Y \setminus \{ 1 \}| \times \phi(p-1)$. 

What's left is to compute the size of $Y$. I can't find a lot to say, but here is a little.
We'll write elements of $S$ as $1+kp$. Such an element is in $X$ if it is of the form $x^p+y^p$ with $x^p + y^p \equiv 1 \mod p$. The latter congruence is equivalent to $x+y \equiv 1 \mod p$. So we have
$$1+kp \equiv (1+z)^p - z^p \bmod p^2$$
$$k \equiv \frac{1}{p} \left( \binom{p}{1} z + \binom{p}{2} z^2 + \cdots + \binom{p}{p-1} z^{p-1} \right) \bmod p.$$
So $Y$ is the set of the distinct values of 
$$f_p(z) := \frac{1}{p} \sum_{j=1}^{p-1} \binom{p}{j} x^j$$
modulo $p$. One can explicitly compute $\frac{1}{p} \binom{p}{j} \equiv \frac{(-1)^{j-1}}{j} \bmod p$ but it doesn't seem to be useful.
$Y$ does contain at least one nonzero value, since $f_p$ only has degree $p-1$, so it can't be identically zero. That answers your request to prove that there is some primitive root of the form $x^p+y^p$. The first few values of $Y$ are
$$\begin{array}{r l}
p & Y \\
3 & \{ 0,2 \} \\
5 & \{0, 1, 2\} \\
7 & \{0, 4, 5\} \\
11 & \{0, 1, 6, 7, 10\} \\
13 & \{0, 5, 6, 7, 9, 10\} \\
\end{array}$$
Here is a plot of $(p, |Y(p)|)$ for the first $100$ odd primes:

The best fit line is $0.40 p + 0.72$. If I were to take a guess, I would guess that $|Y(p)| \approx p/e$ for large $p$. This is because the expected size of the image of a random map $\ZZ/p \to \ZZ/p$ is $\approx p/e$. The numerical fit is so-so, $1/e \approx 0.37$ not $0.40$, but I'm sticking to it until someone gives me a better heuristic.
