Prove that the equation: ${a^k + b^k \equiv c^k}\mod{p}$ has no solutions under the following conditions

Question

Prove that the equation: ${a^k + b^k \equiv c^k}\mod{p}$ has no solutions

where,
$ p $ is a prime $ > 3 $,
$k = \frac{p - 1}{2} $,
and the condition $ 0 < a, b, c < p$ holds.

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See my follow up question here 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

Answer 1

For any natural number which satisfies $(a,p)=1$ with prime $p$, from Fermat's theorem we have $a^{p-1}\equiv 1\pmod{p}$. As, $0<a,b,c<p$, $(a,p)=(b,p)=(c,p)=1$. So all of $a^{\frac{p-1}{2}},b^{\frac{p-1}{2}},c^{\frac{p-1}{2}}$ will satisfy the equation $x^2-1\equiv 0\pmod{p}$. Hence, $x\equiv \pm 1\pmod{p}$. Checking all $8$ cases the result easily follows.