<s>[INST] 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?
 [/INST] 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.
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