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I've searched this website and while there are a few questions similar to mine, I couldn't find what I was looking for/a specific method for what I want to do.

I want to understand how one would prove that the remainder of $5^{336}$ by $23$ is $8$, or in other words, $$5^{336}\equiv 8 \pmod{23}.$$

Can anyone help me understand how one would approach a problem like this?

Martin Sleziak
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HolyThunder
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  • Hint: Look at $5^1, 5^2, 5^3,\ldots$ modulo $23$, and look for a pattern. – vadim123 Feb 02 '15 at 20:42
  • @vadim123 I started out doing so, but I was looking for a more abrangent method. I know that works, but brute-forcing has it's limitations (when no computer/calculator is around) and it takes too long for numbers with longer patterns. – HolyThunder Feb 02 '15 at 20:47
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    @HolyThunder Do you have little fermat / euler at your disposal? – AlexR Feb 02 '15 at 20:47
  • @AlexR now I do. This is a question I came across while studying for an exam but unfortunately I missed about half the semester's classes due to being in the hospital and I missed a lot of important material including the aforementioned Fermat's Little theorem and Euler's. – HolyThunder Feb 02 '15 at 20:54
  • This question might be interesting for you. (And maybe also some other questions in the frequent tab of the tag modular arithmetic.) – Martin Sleziak Feb 03 '15 at 14:59

2 Answers2

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$23$ is prime, so $\phi(23) = 22$. You can thus always take the exponent modulo $22$ (by Euler's theorem, or in this special case Fermat's little theorem). $$5^{336} \equiv 5^6 \equiv 2^3 = 8 \pmod{23}$$ The last congruence is because $5^2 = 25 \equiv 2$

Remark on Brute-Forcing
We actually have $\mathrm{ord}_{23}(5) = 22$ so you'd have to do many computations until you arrive there. This is highly impractical.

Remark on modulus in exponents
We know that $$a^{\mathrm{ord}_n(a)}\equiv 1 \pmod n$$ by definition of the order of an element. Also we know that the order of an element is a divisor of the group order, alas $\mathrm{ord}_n(a) | \phi(n)$. This allows us to make two statements: $$a^k \equiv a^{k\bmod \mathrm{ord}_n(a)} \pmod n$$ and as a corollary of that, $a^k \equiv a^{k\bmod \phi(n)} \pmod n$.

AlexR
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  • Can I always take the exponent modulo $\phi(mod)$? Thanks a lot man, I knew there was a way to solve these without bruteforcing them. – HolyThunder Feb 02 '15 at 20:52
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    @HolyThunder Yes, as you see in the article about Euler's theorem $$a^{\phi(n)} \equiv 1 \pmod{n} \qquad\forall a$$ Computing $\phi(n)$ is easy if you know the prime factorisation of $n$. – AlexR Feb 02 '15 at 20:53
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    Calculating $5^{336}$ directly is not quite impractical: ten multiplications, by my count, as $((((((5^2)^25)^2)^25)^2)^2)^2)^2$, all done modulo $23$. – Lubin Feb 02 '15 at 21:00
  • @Lubin Square-and-multiply reduces the overhead, but that's still bad when doing by hand as opposed to $1$ modulo and $1$ multiplication (don't forget all the moduli in SQM). – AlexR Feb 02 '15 at 21:02
  • Absolutely: your suggestion is the only wise route to the answer. Actually, writing down all powers of $5$ modulo $23$ is not out of the question at all, especially since $5^2=2$. Once you see that $5^{22}=1$, you’re almost home free. – Lubin Feb 02 '15 at 21:02
  • @HolyThunder I've added another remark on taking modulus in exponents. – AlexR Feb 04 '15 at 19:43
  • Ah, cheers. This greatly improves my understanding of modular arithmetics, particularly with exponents. I really need to brush up on these. Thanks! – HolyThunder Feb 04 '15 at 19:52
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Hint $\ {\rm mod}\ 23\!:\ \underbrace{\color{#c00}{5^{\large 22}}\equiv\color{#c00}1}_{\rm little\ Fermat}\Rightarrow\, 5^{\large 336}\equiv 5^{\large 6+\color{#c00}{22}(15)}\equiv \color{#0a0}{5}^{\large \color{#0a0}{2}\cdot 3}\,\color{#c00}{(5^{\large 22})}^{\large 15}\equiv \color{#0a0}2^{\large 3} \color{#c00}{(1)}^{\large 15}\equiv 8 $

where we have used the fundamental Congruence Product and Power Rules.

Remark $ $ Generally, if $\,a^e\equiv 1\pmod m\,$ then exponents on $\,a\,$ can be considered mod $\,e,\,$ i.e. $\ a^{\large j}\equiv a^{\large k}\pmod m\,\ $ if $\,\ j\equiv k\pmod e.\ $ This may be proved exactly as above, i.e.

$$ \begin{array}{}\color{#c00}{a^{\large e}\equiv 1}\\ j = k\! +\! en\end{array}\Rightarrow\,\ a^{\large j}\equiv a^{\large k+en}\equiv a^{\large k}\color{#c00}{(a^{\large e})}^{\large n}\equiv a^{\large k}\color{#c00}{(1)}^{\large n}\equiv a^k\!\!\pmod m\qquad $$

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Bill Dubuque
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