If $5^{2015} \equiv n (\bmod 11) $ and $n \in \{0,1,2,...,10\}$ then what is the value of n?

I have succeeded to solve this problem using Fermat's little theorem and the value of $n$ is $1$ but my problem is different using a theorem or formula I get $ n^2 \equiv 1 (\bmod 11)$....(A)

The theorem is

$n$ is a positive integer and $\gcd(a,n)=1$.If $x^k \equiv a (\mod n)$ and $gcd(k,\phi (n))=d$; $n$ is prime iff $a^{\phi (n)/d} \equiv 1 (\mod n)$ , where $\phi (n)$ is Euler phi function.

Now in the equation (A) ,if we put $n=10$ and $n=1$ both values are satisfied but it is impossible since n must be any one number between $0$ to $10$.

Now I don't know know how to solve the theorem.WHAT IS THE MISTAKE IN THIS METHOD?

  • $\begingroup$ I do not think that the second theorem is valid. Do you perhaps mean Euler's theorem : $$\gcd(a,n)=1\implies a^{\varphi(n)}\equiv 1\mod n$$ ? $\endgroup$ – Peter Jun 23 '20 at 11:14
  • $\begingroup$ I don't know whether it'll satisfy you but to solved this problem i used that $5^5 \equiv 1 mod 11$ and using modular multiplication get n. $\endgroup$ – Szymon Pawlus Jun 23 '20 at 11:14
  • $\begingroup$ No mistake. "$n = 1$ or $n = 10$" is a true statement, since we know $n=1$. "If $n=10$, then $11$ is prime" is a true statement, since the conclusion is true despite the false premise. It's just not quite enough of a conclusion to entirely solve the problem. $\endgroup$ – aschepler Jun 23 '20 at 11:16
  • $\begingroup$ If I'm not wrong, fermat's little theorem is $a^{p-1}\equiv 1 \pmod p$, which I think doesn't give n=1 $\endgroup$ – UmbQbify Jun 23 '20 at 11:23
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    $\begingroup$ @user675453 no obviously ,it does not give the answer in one step.I have done it using FLT and many others theorems. $\endgroup$ – LAMDA Jun 23 '20 at 11:27

Choose $k=2015$ and $x=5$, $gcd(k, \phi(11))=gcd(2015, 10)=5$, then we conclude that $n^\frac{10}{5}=n^2 \equiv 1 \pmod{11}$

However, it doesn't mean that all the solution of $n^2 \equiv 1 \pmod{11}$ is the solution to the original problem.

It just claims that the solution of the original system satisfy $n^2 \equiv 1 \pmod{11}$ since $11$ is a prime.

  • $\begingroup$ but all the steps are equivalent so solution of the last system should be solution of the first and vice versa. $\endgroup$ – LAMDA Jun 23 '20 at 11:32
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    $\begingroup$ If I substitute those numbers in, the theorem reduces to if $5^{2015}\equiv n \pmod{11}$, then $11$ is a prime iff $n^2 \equiv 1 \pmod{11}$. We know that $11$ is a prime. Hence the statement is if $5^{2015}\equiv n \pmod{11}$, then $n^2 \equiv 1 \pmod{11}$. The theorem is of the form of if A, then B iff C. It doesn't claim $C$ implies $A$. $\endgroup$ – Siong Thye Goh Jun 23 '20 at 11:36

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