Calculate $442^{260} \bmod{616}$ using Euler's theorem

Question

I'm asked to calculate $442^{260} \bmod{616}$, using Euler's theorem.
But if I am to use Euler's theorem to solve an expression on the form $a^x \bmod{n}$, then $a$ and $n$ needs to be relatively prime. How can I change the expression above so that I'll end up with an $a$ and a $n$ such that $\gcd(a,n) = 1$?
I was thinking about using the Chinese Remainder Theorem to split up the expression, but since the prime factorization of $616$ is $2^3 \cdot 7 \cdot 11$, I'll always end up with an even $n$ in one of the equations.

Answer 1

Break the modulus into its prime power factors: $$442^{260}\equiv a\bmod8$$ $$442^{260}\equiv b\bmod7$$ $$442^{260}\equiv c\bmod11$$ $a$ is obviously zero as $442^{260}=8\cdot221^3\cdot442^{257}$. Since 442 is relatively prime to 7 and 11, Euler's theorem (actually Fermat's little theorem, as 7 and 11 are prime) can be used for $b$ and $c$: $$442^6\equiv1\bmod7$$ $$442^{260}\equiv442^{6\cdot43+2}\equiv442^2\bmod7$$ Since $442\equiv1\bmod7$: $$b\equiv442^2\equiv1^2\equiv1\bmod7$$ Meanwhile: $$442^{10}\equiv1\bmod11$$ $$c\equiv442^{260}\equiv442^{10\cdot26}\equiv1\bmod11$$ Hence we have $$442^{260}\equiv 0\bmod8$$ $$442^{260}\equiv 1\bmod7$$ $$442^{260}\equiv 1\bmod11$$ and using the Chinese remainder theorem we find that $$442^{260}\equiv1+3\cdot77\equiv232\bmod616$$

Answer 2

Use the Chinese remainder theorem: $\;616=8\times7\times 11$, so \begin{align}\mathbf Z/616\mathbf Z&\simeq\mathbf Z/8\mathbf Z\times\mathbf Z/7\mathbf Z\times\mathbf Z11\mathbf Z\\ 442&\mapsto ([442]_8,[442]_7,[442]_{11})=(2,1,2)\\ 442^{260}&\mapsto (2^{260},1^{260},2^{260})=(0,1,1) \end{align} (We used Lil' Fermat for $2^{260}\bmod11$).

There remains to solve the system of congruences: $$x\equiv\begin{cases}0\mod8,\\1\mod 7,\\1\mod11.\end{cases}$$ We'll begin with the last two congruences. Clearly the solution is $\;x\equiv1\mod 77$.

Next we solve the system $\;x\equiv\begin{cases}0\mod8,\\1\mod 77.\end{cases}$

We need a Bézout's relation between $8$ and $77$. The extended Euclidean algorithm instantly yields $$29\times 8-3\times77=1,$$ so the solution is $$x\equiv 1\times29\times8-0\times3\times77=\color{red}{232}\mod616.$$

Answer 3

Since $\gcd(442,77)=1$ and $\varphi(77)=\varphi(7)\varphi(11)=6\cdot10=60$ we have $442^{60}\equiv1\pmod{77}$ by Euler; Hence $442^{240}=\left(442^{60}\right)^4\equiv1\pmod{77}$.

Now $221\equiv-10\pmod{77}$
$\implies\ 221^3\equiv-1000\equiv1\pmod{77}\ (\because1001=13\cdot77)$;
$\therefore\ 221^{20}\equiv221^2\equiv100\equiv23\pmod{77}$.

Also $2^8=256\equiv25\pmod{77}$
$\implies\ 2^{16}\equiv625\equiv9\pmod{77}$
$\implies\ 2^{17}\equiv18\pmod{77}$. Hence $$2^{17}\cdot221^{20}\cdot442^{240}\equiv18\cdot23\cdot1=414\equiv29\pmod{77}$$ and multiplying through by $2^3=8$ gives $$442^{260}\equiv8\cdot29\ =\ 232\pmod{8\cdot77=616}$$