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I need to show that it doesn't exist any $k \in \mathbb{Z}$ that satisfies $41 \mid (k^5-2)$. Can I apply Fermat's Little Theorem to prove that in anyway? How?

The only thing that came close when I was thinking was that $$41 \mid (k^5-1)(k^5+1)(k^{10}+1)(k^{20}+1)$$ (because $41\mid (k^{40}-1)$ by Fermat's Little Theorem), but from that I couldn't conclude anything. Thanks.

Bill Dubuque
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Gea5th
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  • Follows immediately by applying Euler's Criterion as in the dupe, i.e. raising $,k^5\equiv 2,$ to power $,8,$ contradicts little Fermat. Or raise it to power $,4,$ to get $,k^{20}\equiv 16\not \equiv \pm1,$ so $,k^{40}\not\equiv 1$ by $\bmod p!: x^2\equiv 1\Rightarrow (x-1)(x+1)\equiv 0\Rightarrow x\equiv \pm1,$ by $p$ prime. So it is redundant to check that $,x = k^4\equiv 2,$ is not a root of some factor of $,x^8-1 = (x-1)(x+1)(x^2+1)(x^4+1)$ since it suffices to verify $,x^4\not\equiv \pm1\ \ $ – Bill Dubuque May 01 '21 at 23:42

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Using what you tried, first note $41 \mid (k^5 - 2) \implies k^5 \equiv 2 \pmod{41}$. Also, for any such $k$, then $41$ (being prime) must divide at least one of $k^5 - 1$, $k^5 + 1$, $k^{10} + 1$ and $k^{20} + 1$. However, $k^5 - 1 \equiv 2 - 1 = 1 \pmod{41}$, $k^5 + 1 \equiv 2 + 1 \equiv 3 \pmod{41}$, $k^{10} + 1 \equiv (2)^2 + 1 \equiv 5 \pmod{41}$ and $k^{20} + 1 \equiv 2^{4} + 1 \equiv 17 \pmod{41}$. Thus, since $41$ doesn't divide any of those factors, there can't be any such $k$.

Alternatively, assume there's a $k \in \mathbb{Z}$ such that $41 \mid (k^5 - 2)$. Then this also means that

$$k^{40} = (k^5)^8 \equiv 2^8 = 256 \equiv 10 \pmod{41} \tag{1}\label{eq1A}$$

However, since $41 \not\mid k$ and $41$ is prime, then Fermat's little theorem states that $k^{40} \equiv 1 \pmod{41}$, which contradicts \eqref{eq1A}. Thus, there is no such $k$.

John Omielan
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