Showing there is no triplet of positive integers $(a,b,c)$ satisfying $a^7+b^7=7^c$ 
Show that $$a^7+b^7=7^c$$ has no positive integer solutions $(a,b,c)$.

I've posted a general and way too long approach as an answer. How may one prove the claim more briefly and specifically?
 A: If $(a_0,b_0,c_0)$ is the solution then $(7^xa_0,7^xb_0,c_0+7x)$ is also a solution. Therefore, without loss of generality, assume that $7 \nmid a, 7 \nmid b$.
By applying Fermat's little theorem we have $a^7+b^7 \equiv a+b \pmod 7$. Therefore $7|a+b$. We then apply Lifting The Exponent lemma, we get $v_7(a^7+b^7)=v_7(a+b)+v_7(7)=c$ or $v_7(a+b)=c-1$. Let $a+b=7^{c-1} \cdot k$ with $k \in \mathbb{Z^+}, 7 \nmid k$. The equation is equivalent to $$k \cdot \frac{a^7+b^7}{a+b}=7.$$
WLOG $a \ge b$. Note that if $a \ge 2$ then $\frac{a^7+b^7}{a+b} > a+b$. From this, we have $7^{c-1} \cdot k^2 <7$. Hence, $c=1$. We get $a^7+b^7=7$. This equation has no solution. 
Therefore, $a \le 1$ or $a=b=1$. We obtain $7^c=2$, no solution.
Thus, the equation has no solution in positive integers.
A: More generally, $a^p+b^p=p^c$ has no solutions ($a,b,c\ge 1$, $p$ prime), except for $$(\{a,b\},c,p)=(\{2,1\},2,3),(\{2^k,2^k\},2k+1,2),\,k\in\Bbb Z_{\ge 0}$$


*

*$a=b$. Then $(a,b,c,p)=(2^k,2^k,2k+1,2)$, $k\in\Bbb Z_{\ge 0}$.

*$p=2$. Let $(a,b)=(2^ka_1,2^kb_1)$, $k\ge 0$ and $a_1,b_1\ge 1$ odd. $a_1^2+b_1^2=2^{c-2k}$.
$4\mid a_1^2+b_1^2\,\Rightarrow\, 2\mid a_1,b_1$, so $c-2k\in\{0,1\}$, $(a,b,c,p)=(2^k,2^k,2k+1,2)$.

*wlog $a>b$, $p>2$. $(a,b,c,p)=(2,1,2,3)$, otherwise by Zsigmondy's theorem $$a^p+b^p=(a+b)(a^{p-1}-a^{p-2}b+\cdots+b^{p-1})$$ has a prime divisor that does not divide $a+b\ge 2$, so has at least two prime divisors.

