<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] Very nice! Straightforward and simple.
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