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I want to determine the form (or function) of the natural numbers $x$ such that $$5^x \equiv 1 \pmod{13}$$ I know how to do that with polynomial forms but I haven't seen an example where $x$ is in exponential form.
if we take the first terms of $x$, $x_1=0$, $x_2=1$ etc. we have for $x_1$ $$5^0 \equiv 1 \pmod{13}$$ $$1 \equiv 1 \pmod{13}$$ for $x_2 = 1$ $$5^1 \equiv 1 \pmod{13}$$ $$5 \equiv 1 \pmod{13}$$ I don't understand how we can conclude the form of $x$ by finding some first terms and eventually by induction

William
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    This is not clear. $5^1$ is not $1\pmod {13}$. – lulu Apr 20 '21 at 17:53
  • The first step is to form a guess as to which $n$ satisfy $5^n\equiv 1 \pmod {13}$. Then use induction to prove it. – lulu Apr 20 '21 at 17:54
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    You are supposed to find a pattern. One can hardly spot a pattern with only two samples. – jjagmath Apr 20 '21 at 17:54
  • Note $5^2\equiv-1\bmod13$ – J. W. Tanner Apr 20 '21 at 18:30
  • Compute further and look for periodicity in the sequence $,a_n := 5^n\bmod 13,,$ then prove by induction that the observed pattern persists for all $n\in \Bbb N.,$ This cyclicity will be clarified when you study the multiplicative order of elements – Bill Dubuque Apr 20 '21 at 19:34

1 Answers1

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Fermat's Little Theorem tells us that, for any natural number $a$ and any prime number $p$

$a^{p-1} \equiv 1 \pmod {p}$

So if

$x \equiv 0 \pmod {12}$ $\to$ $5^x \equiv 1 \pmod {13}$

Once we have found a periodic solution we can check for values of x up to 12 and, if there is no solution there, there won't be any in the other repetitions of the period. For example, once we have checked that

$5^3 \not\equiv 1 \pmod {13}$

we know that

$x\not\equiv 3 \pmod {12}$

After checking all values for x less than $12$ i see that the congruence holds for $x=0$, $x=4$, $x=8$ and of course $x=12$. Wait, there is a pattern! The period is not $12$, it is $4$, so all solutions are

$x \equiv 0 \pmod 4$

This is a fast strategy to find solutions, but it does not provide a perfectly rigorous proof of the fact that the ones we have found are the only solutions. We can prove this result by showing that

$5^x \equiv 1 \pmod {13}$ $\to$ $5^x\times5^4 \equiv 5^4 \equiv 1 \pmod {13}$

So if $5^x \equiv 1 \pmod {13}$, than $5^{x+4} \equiv 1 \pmod {13}$

Otherwise, if $5^x \not\equiv 1 \pmod {13}$, than $5^{x+4} \not\equiv 1 \pmod {13}$

This means that if we start with a number $a$ in the interval $[0,1,2,3]$ that solves the congruence, all values $4k+a$ will be solutions, otherwise if $a$ is not a solution, $4k+a$ won't be either. So since in the interval [0,1,2,3] the only solution is $0$, all the solution to this problem are in the form

$x=4k$

Claudio_DiBiase
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