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To determine this for $13^{33} \pmod{64}$ is easy since $\phi(64)=32$ and $\gcd(64,33)=1$, we have $13^{\phi(64)}=13^{32}\equiv1 \pmod{64}$. This means that

$$13^{33}=13^{31}\cdot13\equiv 13 \equiv1 \pmod{64}$$

But this method don't apply to $3^{18} \pmod{37}$ since $\phi(37)=36>18.$ How can I do this?

rtybase
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Parseval
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    Hint: $\ 36 = 2\cdot 18.\ $ Do you know Euler's criterion or quadratic reciprocity or related ideas? – Bill Dubuque Mar 10 '19 at 23:28
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    Maybe this will help $3^{36} = 3^{18} 3^{18} \equiv 1 \mod 37$ – JoseSquare Mar 10 '19 at 23:30
  • @BillDubuque - Id much appreciate if you explain those for me if you have time. I have another problem I i might apply it on now, a problem which the suggestions above will not work: $2^{100} \pmod{257}.$ – Parseval Mar 10 '19 at 23:40
  • @Parseval If you don't know them then you were probably meant to use another method. I added a simple way in my answer, by using that $3$ is a square. For the general idea see Euler's Criterion. – Bill Dubuque Mar 11 '19 at 00:39
  • @BillDubuque: I am only watching the front page, but as it says the linked question is a master question / abstract duplicate for every question of the form "how do I compute $a^b\pmod{c}$. If there isn't an answer there that covers this, that is the best place to put one. – Carl Mummert Mar 11 '19 at 01:29
  • Not a dupe of that generic thread unless you can specifically point to answers that handle all the possible ways of solution here. – Bill Dubuque Mar 11 '19 at 01:42
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    @BillDubuque Not only does that dupe target describe techniques useful for this thread, the 150 (!) threads linked to it surely cover all the missing holes. I consider your action blatant abuse of the dupehammer privilege. – Jyrki Lahtonen Mar 11 '19 at 17:04
  • @Jyrki I disagree. – Bill Dubuque Mar 11 '19 at 17:08
  • Dupe closures should not be abused to send askers on possible wild goose chases (that may or may not be successful). That generic thread is a motley mix of unrelated techniques. It is far from complete, and (pedagogical) quality leaves much to be desired. If one has the (extensive) number theory background required to know for sure that some answer(s) there cover all the ways one would attack some question then one should specifically link to said answers (and elaborate on how they apply if need be). – Bill Dubuque Mar 11 '19 at 18:01
  • I see your point @BillDubuque. I rather feel that the burden of finding the best duplicate match should rest on the shoulders of those who ask and those who answer. Not necessarily on the crew of janitors. I guess there is room for differences of opinions here. – Jyrki Lahtonen Mar 12 '19 at 05:47

4 Answers4

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Hint. From Fermat's little theorem, given $37$ is prime and $\gcd(3,37)=1$ $$3^{36} \equiv 1 \pmod{37}$$ or $$37 \mid \left(3^{18}+1\right)\cdot \left(3^{18}-1\right)$$ According to Euclid's lemma one of these should happen $37 \mid \left(3^{18}+1\right)$ or $37 \mid \left(3^{18}-1\right)$

But, there is an "easier" way $$27\equiv 3^3 \equiv -10 \pmod{37} \Rightarrow \\ 3^{18} \equiv 10^6 \equiv 100^3 \pmod{37} \Rightarrow \\ 3^{18} \equiv {-11}^3 \equiv 121\cdot(-11)\pmod{37} \Rightarrow \\ 3^{18} \equiv 10\cdot(-11) \equiv - 110\pmod{37} \Rightarrow \\ 3^{18} \equiv 1\pmod{37}$$ because $37 \mid 111$ or $100 \equiv -11 \pmod{37}$.

rtybase
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$\bmod 37\!:\,\ 1\equiv 5(15)\,\overset{\large \times 3}\Longrightarrow\, 3\equiv 15^{\large 2}\,\overset{(\ \ )^{\Large 18}}\Longrightarrow\, 3^{\large 18}\!\equiv 15^{\large 36}\!\equiv 1\ $ by little Fermat

Bill Dubuque
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mod $37,$ $ \color{blue}{3^4=81\equiv 7},$ so $ 3^8\equiv (\color {blue }{3^4})^2 \equiv \color{blue}7^2=49 \equiv 12, $ so $\color{red}{3^9\equiv 3\times}$$\color{black}{3^8}\color{red}{\equiv 3 \times} \color{black}{12} \color{red}{\equiv 36 \equiv -1},$

so $ \color{green}{3^{18}\equiv (3^9)^2\equiv 1 }$

J. W. Tanner
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Applying the quadratic reciprocity theorem, there is no need to calculate much:

$3$ is a square $\mod 37$, iff $37$ is a square $\mod 3$. (Note: $37 = 4n+1$).

Since $37 \equiv 1 \mod 3$, is a square, then: $3$ is a square $\mod 37$.

All squares power to $18$ are $1 \mod 37$. (Note: $18 = (37-1) / 2$).

And all not squares power to $18$ are $-1\equiv36 \mod 37$

Because $3$ is a square, then $3^{18} \equiv 1 \mod 37$

With this same argument it is proved that: $5 ^ {18} \equiv -1 \mod 37$. Since $5$ is not a square $\mod 37$. Since $37 \equiv 2 \mod 5$ is not a square $\mod 5$.

Angel Moreno
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  • Op does not know QR - see their response when I mentioned it in comments on the question. That's why I explicitly showed $3$ is a square in my answer. – Bill Dubuque Mar 11 '19 at 21:26