Prove that $2^{6ℓ + 2}≡ 4 \pmod{18}$ for $0\leℓ ∈ ℤ$.
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1Do you mean $2^{6\ell+2}$ or $2^6\ell+2$ or even $2^{6\ell}+2$? – Bernard Dec 21 '20 at 18:05
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1Welcome to MSE. Please note that askers are expected to provide context for their questions, as is explained here. For example, what are your thoughts on the problem? What have you tried so far? Where did you encounter this question? – Ben Grossmann Dec 21 '20 at 18:06
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@Bernard I think it is $2^{6l+2}$ because in the body it is written 2^(6l+2) – Dec 21 '20 at 18:09
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You can use the same methods as in the linked dupe. – Bill Dubuque Dec 21 '20 at 18:19
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1Welcome to Mathematics Stack Exchange. Since you tagged [tag:totient-function], note that $\phi(9)=6$, so $2^{6\ell}\equiv1\pmod9$, so $9|2^{6\ell+1}-2$, so $2^{6\ell+2}\equiv4\bmod18$ – J. W. Tanner Dec 21 '20 at 18:21
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More generally, $a^{6+2} \equiv a^2 \bmod{18}$ for all $a \in \mathbb Z$. See Wikipedia – lhf Dec 21 '20 at 18:32
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How can I use Euler function in here? – אבג דיזיין Dec 21 '20 at 18:39
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ברוך הבא. See the above comments – J. W. Tanner Dec 21 '20 at 19:14
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TBH can't really see how you moved from the second "so" to the last one.. @J.W.Tanner – אבג דיזיין Dec 21 '20 at 19:17
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perhaps it would have been more clear had I written $2^{6\ell+1}\equiv2\bmod9$ instead of the equivalent $9|2^{6\ell+1}-2$ – J. W. Tanner Dec 21 '20 at 19:19
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@J.W.Tanner It is okay, but I just can't see how you made the move in the end from mod 9 to mod 18. – אבג דיזיין Dec 21 '20 at 19:21
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Could you see it if I write $9|2^{6\ell+1}-2$, so $18|2^{6\ell+2}-4$? – J. W. Tanner Dec 21 '20 at 19:22
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@J.W.Tanner Mabey you can do me a favor and write down the full answer. I believe your answer is correct but I'm having a hard time understanding the moves. – אבג דיזיין Dec 21 '20 at 19:25
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I can't post an answer now, because the question has been closed as a duplicate, though I'm not sure my answer would apply to the proposed duplicate; anyway, it might help you to look at this – J. W. Tanner Dec 21 '20 at 19:34
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Hint: This follows from the fact that $$ 2^6 \cdot 4 \equiv 4 \pmod{18}. $$
Ben Grossmann
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More generally, $a^{6+2} \equiv a^2 \bmod{18}$ for all $a \in \mathbb Z$. See Wikipedia – lhf Dec 21 '20 at 18:31