<s>[INST] $5^{th}$ power of any integer is of the form $11k$ or $11k +1$ or $11k -1$. $5^{th}$ power of any integer is of the form $11k$ or $11k +1$ or $11k -1$.
Let the integer be $x$. If $x$ has a factor $11$ then $x^5$ is of the form $11k$. Now we consider the case where $11 \not\mid x$ .
By Fermat's Theorem we know that $x^{10} \equiv 1(\mod 11)$. Thus $11 \mid x^{10} -1$ i.e. $11 \mid (x^{5} -1)(x^5 +1)$. Since $11$ is prime, $11 \mid (x^{5} -1)$ or $11 \mid (x^{5} +1)$. Thus $x^5$ is of the form $11k +1$ or $11k -1$.
Is the proof correct?
 [/INST] Here is an alternative proof (though I find yours nicer):


*

*$x\equiv 0\pmod{11}\implies x^5\equiv 0^5\equiv11\cdot   0  \equiv 0\pmod{11}$

*$x\equiv 1\pmod{11}\implies x^5\equiv 1^5\equiv11\cdot   0+1\equiv+1\pmod{11}$

*$x\equiv 2\pmod{11}\implies x^5\equiv 2^5\equiv11\cdot   3-1\equiv-1\pmod{11}$

*$x\equiv 3\pmod{11}\implies x^5\equiv 3^5\equiv11\cdot  22+1\equiv+1\pmod{11}$

*$x\equiv 4\pmod{11}\implies x^5\equiv 4^5\equiv11\cdot  93+1\equiv+1\pmod{11}$

*$x\equiv 5\pmod{11}\implies x^5\equiv 5^5\equiv11\cdot 284+1\equiv+1\pmod{11}$

*$x\equiv 6\pmod{11}\implies x^5\equiv 6^5\equiv11\cdot 707-1\equiv-1\pmod{11}$

*$x\equiv 7\pmod{11}\implies x^5\equiv 7^5\equiv11\cdot1528-1\equiv-1\pmod{11}$

*$x\equiv 8\pmod{11}\implies x^5\equiv 8^5\equiv11\cdot2979-1\equiv-1\pmod{11}$

*$x\equiv 9\pmod{11}\implies x^5\equiv 9^5\equiv11\cdot5368+1\equiv+1\pmod{11}$

*$x\equiv10\pmod{11}\implies x^5\equiv10^5\equiv11\cdot9091-1\equiv-1\pmod{11}$

</s>