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

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

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

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    $\begingroup$ Hint: let $u = a^k, v = b^k, w = c^k$, we have $u^2 \equiv v^2 \equiv w^2 \equiv 1 \pmod p$. Now $4u^2v^2 \equiv (w^2 - u^2 - v^2)^2 \pmod p \implies \cdots$ $\endgroup$ – achille hui Jul 24 '18 at 12:36
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    $\begingroup$ I like the start of achille hui's hint. It's pretty simple after the first step. What are the possibilities? $\endgroup$ – John Brevik Jul 24 '18 at 12:50
  • $\begingroup$ $a^{\frac{p-1}{2}}\pmod{p}$ is the Legendre symbol $\left(\frac{a}{p}\right)$. $\endgroup$ – Jack D'Aurizio Jul 24 '18 at 13:21
  • $\begingroup$ Thanks for the hint @achillehui . Now it can be proved using Fermat's theorem as in tarit's answer. $\endgroup$ – Rahul Goswami Jul 24 '18 at 13:26
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    $\begingroup$ For the values of $k$ coprime with $p-1$, solutions always exist: in fact, what it happens is that any integer is a $k$-power. For k dividing $p-1$, it will depend on $p$ and $k$.... $\endgroup$ – xarles Jul 24 '18 at 14:25

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.


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