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The answers says

"According to Fermat's little theorem, $12^{16} \equiv 1 \pmod{17}$."

My question: Why is $12^{16}$ interesting? How do we get $12^{16}$ from $12^{12^{12}}$?

The answers then say

"Therefore we calculate $12^{12} = 144^{6} \equiv 0^6 \pmod{16} = 0$.

My question: Why $12^{12}$? Why $\mod{16}$?

Shaun
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saner
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    Do you know Fermat's Little Theorem? Do you know how to find the residue of $12^{12}$ modulo $16$? Readers will be more responsive if you fill in some details you do understand. – hardmath Dec 18 '21 at 16:49
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    See mod order reduction in the linked dupe, which explains how to reduce modular powers mod any known period (cycle), e.g. the least length period (= order). See the linked posts there for many worked examples. Above, by lil Fermat, $16$ is a known period of $12^x\pmod{!17},$ (i.e. $,12^{16}\equiv 1\pmod{17},$ so $12^n\equiv 12^{n\bmod 16},,$ Yours is case $n = 12^{12}\ \ $ – Bill Dubuque Dec 18 '21 at 17:12

1 Answers1

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The sentence

Therefore we calculate $12^{12}=144^6\equiv0^6(mod\ 16)=0$

is simply proving that $12^{12}$ is divisible by $16$.

After it's proven, we can state $12^{12}$ in the form $16k$ where $k$ is a positive integer. So,

$$12^{12^{12}}=12^{16k}=(12^{16})^k\equiv1^k=1(mod\ 17)$$

user_194421
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  • Thanks. My provided answer citation didn't make it clear if they are talking about the (12^12) in the "bottom" or on "top" and your answer made it clear. – saner Dec 18 '21 at 19:45