Check whether $k \in [0, p]$ in the equation: ${a^k + b^k \equiv c^k}\mod{p}$ has no solutions under the following conditions Check whether the equation:  ${a^k + b^k \equiv c^k}\mod{p}$ has no solutions
where,
$ p $ is a prime $ > 3 $,
$k \in [0, p]$
and the condition $ 0 < a, b, c < p$ holds.  
Can we determine all values of $k \in [1, p] $ for which solution doesn't exist?
Or I rephrase - for any $k \in [1, p]$, how to tell if any solution exists or not? Thanks. 
Some Observations:
We are considering $p > 3$, so $(p - 1)$ is always even. We know that there is at least one such $k$ for which solution doesn't exist ($\frac{p - 1}{2}$).
Now, there are some interesting observation/pattern in such values of $k$ (for which there is no solution).  


*

*If a solution doesn't exist for some $k \in [0, p-1)$ , then there is no solution for all $k + n \dot (p - 1), n \in W$.

*If $\frac{p - 1}{2} \in Prime$, then there in only one such $k$ i.e. $\frac{p-1}{2}$.

*If $\frac{p-1}{2} \notin Prime$, then there are/maybe other such $k$. Now, if there is such $k < \frac{p-1}{2}$ for which there is no solution, then $k | (p - 1)$, i.e. $p-1$ is divisible by $k$. In other words, if $k$ and $p-1$ are co-prime, then a solution always exists.  

*Following from previous point, then all such $\in [k, 2k, 3k, .. (p-1)]$ don't have any solution.

*Some examples,


*

*$p = 11$:
$\frac{p-1}{2} = 5 \in Prime$, so only one such k. $\{5\}$. (Note we can generate all $k$s as $\{0, 5, 10, 15, 20, ...\}$.)

*$p = 13$:
$\frac{p-1}{2} = 6 \notin Prime$, other such $k < \frac{p-1}{2}$ are 3 and 4. All such $k \in [0, p-1]$ are $\{0, 3, 4, 6, 8, 9, 12\}$.

*$p = 47$:
$\frac{p-1}{2} = 23 \in Prime$, so only one such $k = 23$. $\{0, 23, 46\}$.

*$p = 67$:
such $k$s are $\{0, 22, 33, 44, 66\}$

*$p = 29$:
such $k$s are $\{0, 7, 14, 21, 28\}$


 A: Write $k = k'd$ where $k' = gcd(k,p-1)$ is prime to $p-1$; then $x\mapsto x^d$ is a bijection of $\mathbb{F}_p$, and thus writing $X' = X^d, Y'=Y^d, Z'=Z^d$, we see that the equation with $k$ has the same number of solutions as the equation with $k'$. Hence let us assume that $k$ divides $p-1$.
Your question is closely related to counting the number points on the projective Fermat curve $X^k + Y^k = Z^k$ over $\mathbb{F}_p$. That is, we count the solutions for which $(X,Y,Z) \neq (0,0,0)$ and we identify any two solutions of the form $(X,Y,Z)$ and $(\lambda X,\lambda Y,\lambda Z)$ where $\lambda \in \mathbb{F}_p^\times$. More precisely, the number of projective solutions is simply $(p-1)$ times the number of solutions to your equation when $(X,Y,Z) \neq (0,0,0)$ (you also prohibit solutions of the form where one of $X,Y$ or $Z$ vanishes, but it is easy to count how many solutions are there of this form; I will leave this calculation to you).
Let us focus on counting the number of projective solutions then. Let us denote it by $N_p$. This question is very classical, and it is known how to present the number of solutions (when $k$ divides $p-1$) using certain Jacobi sums, but I doubt this will be of much use to you here. However, as a direct consequence of the use of the Jacobi sums, one obtains the very nice Hasse-Weil estimate $$|N_p - (p+1)| \leq (k-1)(k-2)\sqrt{p}$$
In particular, for any given $k$ there are many projective solutions (and hence a solution in your case) for $p$ sufficiently large.
While this does not give a complete answer to your question, note that the complete formula with Jacobi sums will tell you exactly how many solutions there are, but since they are sums of convolutions of characters, I think their behaviour can be quite random. Thus for $k$ large relative to $p$ (say $k\approx p^{1/4}$) I think it will be hard to say something general.
For more about the Jacobi sums and the Fermat curve, see the nice treatment in section 6.1 of the book "Introduction to Cyclotomic Fields" by Lawrence Washington, where all the claims I mentioned here are proved. 
Edit: let me be a bit more precise. Let us write the number of solutions for your equation as $M_p$. You are asking when $M_p > 0$. We have the easy estimate $$M_p \geq (p-1)(N_p - 3k),$$ where the $-3k$ comes from the need to disregard solutions of the form $(X,Y,0),(X,0,Z),(0,Y,Z)$. So whenever $N_p >3k$, there is a solution. When does this happen? Well, using the Hasse-Weil bound, we have $$N_p -3k \geq p+1 -3k -(k-1)(k-2)\sqrt{p}.$$ If my high school algebra does not fail me, this inequality is achieved whenever (say) $k < \frac{1}{2}p^{1/4}$, at least if $p>10$. You may check it for yourself (one should also be able to achieve a slightly better bound; in any case the best asymptotic bound is $k<< p^{1/4}$). 
Of course, in general one has to replace $k$ by $gcd(k,p-1)$. Therefore, this method shows that you have a solution whenever $gcd(k,p-1) <  \frac{1}{2}p^{1/4}$ (which is in most cases a much weaker inequality than $k < \frac{1}{2}p^{1/4}$).
