Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$

Question

I ran into this next question:

Show that if $a$ is coprime to $210$:

$a^{12}\equiv1 \pmod{210}$.

This is not a homework question.

Thank you very much in advance,

Yaron

Answer 1

Hint: Since $210$ is the product of the pairwise relatively prime numbers $2$, $3$, $5$, and $7$, it is enough to show that (i) $a^{12}\equiv 1\pmod{2}$, and (ii) $a^{12}\equiv 1\pmod{3}$, and (iii) $a^{12}\equiv 1\pmod {5}$, and (iv) $a^{12}\equiv 1\pmod{7}$.

Each of these four tasks can be done using Fermat's Theorem.

Each of the tasks can also be done using somewhat less machinery, but then writing out the details takes quite a bit longer.

Detail: We show for example that $a^{12}\equiv 1\pmod{7}$. We have that $a$ and $7$ are coprime. Fermat's Theorem says that if $p$ is prime, and $a$ and $p$ are coprime, then $a^{p-1}\equiv 1\pmod{p}$. Taking $p=7$, we find that $a^6\equiv 1\pmod{7}$. Squaring both sides of this congruence, we find that $a^{12}\equiv 1\pmod{7}$.

The argument for $5$ is almost the same. We have by Fermat's Theorem that $a^{4}\equiv 1\pmod{5}$. Since $4$ divides $12$, it follows that $a^{12}\equiv 1\pmod{5}$. The same idea works for $3$ and for $2$. In the case of $2$ at least, it is silly to use Fermat's Theorem. For $a$ and $2$ coprime just says that $a$ is odd. And if $a$ is odd, so is $a^{12}$, which says that $a^{12}\equiv 1\pmod{2}$.

Answer 2

HINT:

Using Carmichael Function,

$$\lambda(210)=\text{lcm}\{\lambda(2),\lambda(3),\lambda(5),\lambda(7)\}=\text{lcm}(1,2,4,6)=12$$

In fact,we can find a larger $n$ such that $a^{12}\equiv\pmod n$ for all $a$ co-prime to $n$

We need $\lambda(n)$ divides $12$

Case $1: $ If $p=2,$ we know, $\lambda(2^r)=\begin{cases} 2^{r-1} &\mbox{if } 1\le r \le 2 \\ 2^{r-2} & \text{ otherwise }\end{cases}$

$2^{r-2}$ needs to divide $12,r-2\le 2\implies r \le 4$

Case $2: $ If $p$ is odd prime, $\lambda(p^r)=\phi(p^r)=p^{r-1}(p-1),$

If $r>1,$ as the only odd divisor$(>1)$ of $12$ is $3^1\implies p=3,r=2$

Else $r=1\implies p-1$ must divide $12\implies p$ can be $2,3,5,7,13$

$2,3$ have already been encountered

So, the highest value of $n$ will be $13^1\cdot7^1\cdot5^1\cdot3^2\cdot2^4$

As $a^{12}\equiv1\pmod {13\cdot7\cdot5\cdot3^2\cdot2^4}$ if $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1,$

$a^{12}\equiv1\pmod d$ where $d$ is any divisor of $13\cdot7\cdot5\cdot3^2\cdot2^4$ and $(a,d)=1$ as $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1$

Answer 3

Since you have to consider only $a$ modulo $210$, you is only a finite number of case to check. And since it is not a homework, everybody will trust you if you claim you used your computer.

Answer 4

Hint $\ $ Either apply Carmichael's generalization of Euler-Fermat, or, directly, by little Fermat

$$A^{N_j}\equiv 1\ \ ({\rm mod}\ M_j)\ \Rightarrow\ A^{{\rm lcm}\ N_j}\equiv 1\ \ ({\rm mod\ lcm}\ M_j)$$

$${\rm for}\quad \begin{cases} \:N = (1,2,4,6)\,\Rightarrow\,{\rm lcm}\ N_j\, = 12\\ \, M = (2,3,5,7)\,\Rightarrow\, {\rm lcm}\ M_j = 210\end{cases}\ \ \ \ $$

Answer 5

hints:first find phi(210)=the number of positive integers less than 210 and prime to 210....then apply Euler's theorem : a^phi(210) congruence to 1 mod(210)...