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We want to find the number of items of the set $\{a \in \mathbb{Z}_{24} : a^{1001} = [5]\} $

Case: $\gcd(a,24)=1$

Using

  • Euler's Theorem (if $(\gcd(a,n))=1 \Rightarrow a^{\phi(n)}\equiv1\bmod n$)
  • $\phi(24)=\phi(2^3)\cdot\phi(3)=4\cdot2=8$
  • $1001=8\cdot125+1$

we get,

$$a^{1001}=a^{8\cdot125}\cdot a \equiv a\bmod24 \equiv 5\bmod 24\Leftrightarrow a \equiv 5 \bmod24.$$

Case: $\gcd(a,24)\neq 1$

Any help would be greatly appreciated!

J. W. Tanner
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    Second case can't work: if $d\mid a;$ and $;d\mid 24;$ and $;24\mid a^{1001}-5$, then $d\mid 5,$ so $;d\mid1$ – J. W. Tanner Nov 26 '23 at 17:35
  • $a^{1001}!= 5\Rightarrow a\mid 5,$ so $,5$ unit $\Rightarrow a,$ unit, by divisors of units are units (cf. 2nd dupe). The rest (which you've done) is a special case of mod order reduction using Euler's Theorem, see the first dupe. Recall unit means invertible, which here are the elements coprime to $24.\ \ $ – Bill Dubuque Nov 26 '23 at 18:53
  • More generally a root $,r,$ of a polynomial $,f(x),$ divides its constant coef $,f(0),,$ so $,f(0),$ unit $\Rightarrow r,$ unit, e.g. see inversion via norms, e.g. rationalizing denominators, e.g. in OP $$\dfrac{1}{\color{#c00}a} ,=, \dfrac{a^{1000}}{a^{1001}},=,\dfrac{a^{1000}}{\color{#0a0}5}\qquad$$ shows how we can view $,\color{#0a0}5,$ invertible $,\Rightarrow \color{#c00}a,$ invertible as "rationalizing" the denominator of $\large ,\frac1{\color{#c00}a}\ \ $ – Bill Dubuque Nov 26 '23 at 20:12

1 Answers1

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The comment under the question already provides an answer, but let me give another.

Notice that $\gcd(5, 24) = 1$, so $5$ has multiplicative inverse modulo $24$.

On the other hand, any $a$ such that $\gcd(a, 24) \neq 1$ is not invertible, so any of its powers cannot be invertible as well. Therefore no power of $a$ can give you $5$ modulo $24$.

Esgeriath
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing). – Bill Dubuque Nov 26 '23 at 18:48