Determining whether there exists an integer $a$ such that $\text{ord}_{20}(a) = 8$.

Question

I am trying to determine whether there exists an integer $a$ such that $\text{ord}_{20}(a) = 8$. I know that if $(a,n) = 1$ and $n>0$, then $\text{ord}_{n}(a)\mid \phi(n)$. I cannot use any abstract algebraic techniques. I am only expected to know what primitive roots are, Euler's Theorem, and some other elementary number-theoretic principles related to congruences, linear diophantine equations, and Fermat's Little Theorem.

I understand that I can answer this question by going through each of $1,2,3,\ldots,20$ and checking to see if each raised to the power of $8$ is congruent to $1$ mod 20. However, there is probably a faster way to do this and I am looking for help from the StackExchange community on how to do that. Thank you in advance.

Answer 1

Let $a$ be an integer coprime with $20$. Then $a$ is coprime with both $4$ and $5$.

By the Fermat—Euler theorem, we have $a^2 \equiv 1 \bmod 4$ and $a^4 \equiv 1 \bmod 5$.

By the Chinese Remainder theorem, these congruences imply $a^4 \equiv 1 \bmod 20$.

Hence, there is no integer having order $8$ mod $20$.

Answer 2

No there does not exists any such element. Because if $\exists~ a\in U_{20} $ then since order $U_{20} $ is $\phi(20)=\phi(2^2.5)=8$ which shows that $U_{20} $ is cyclic group. This is a contradiction since we know that only $\displaystyle U_{p^n}~\&~U_{2p^n}$ where $p$ is odd prime , are cyclic groups.