Modular arithmetic with a gigantic exponent (ie. $2^{2^{403}} \equiv x$ (mod $23$))

Asked Nov 13 '17 at 00:05

Active Nov 13 '17 at 00:05

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Suppose I want to find $2^{2^{403}} \equiv x$ (mod $23$). Now I know by Fermat's Little Theorem that $2^{22} \equiv 1 \space (\mod 23)$ and I know how to show that $2^{403} \equiv 8 \space (\mod 22)$. But I don't know how to put these two together to solve the congruence, or if I'm missing some crucial simplification step or theorem. I can't even use WolframAlpha because the number is too big.

My question is: How do you solve $2^{2^{403}} \equiv x$ (mod $23$)?

asked Nov 13 '17 at 00:05

BigSpicyPotato

1

If $2^{403} \equiv 8 \pmod{22}$, then $2^{403} = 22k + 8$ for some integer $k$. So, $2^{2^{403}} \equiv 2^{22k + 8} \equiv (2^{22})^k \cdot 2^{8} \pmod{23}$. Does this lead you somewhere? – Henrique Augusto Souza Nov 13 '17 at 00:08
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You've already done all the hard work. – Randall Nov 13 '17 at 00:09
@HenriqueAugustoSouza that worked thanks. If you wanna put that in an answer i'll mark it as the answer – BigSpicyPotato Nov 13 '17 at 00:11
What is the source of this problem? It, or parts of it, have turned up a lot recently. – lulu Nov 13 '17 at 00:13
@lulu It is a question from a university math course – BigSpicyPotato Nov 13 '17 at 00:24

Modular arithmetic with a gigantic exponent (ie. $2^{2^{403}} \equiv x$ (mod $23$))

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