Find 3 integers a such that $ a\equiv 19^{10562}(mod\: 40320) $

Question

I need to find three integer $ a\equiv 19^{10562}(mod\: 40320) $. I know that I need to use Eulers Formula $a^{\phi (m)}\equiv 1(mod\: m)$. I have determined that $\phi (m)=9216$. So using this I think $19^{9216}\equiv 1(mod\, 40320)$. Now I believe that gets me to the following.

$$19^{10562}\equiv 19^{9216}\cdot 19^{1346}(mod\, 40320)$$

But after this I am unsure where to go with this because I'm not sure how to address the $19^{1346}(mod\, 40320)$. Any insight would be appreciated.

Thanks,

Answer 1

I don't know if I am correct, but let's see... Supposing I am correct, this is very easy.

Notice that $19^{10562}$ itself would be congruent to itself in mod 40320. So will $19^{10562}+40320$, and so will all numbers $19^{10562}+40320n$ where $n$ is an integer...

Answer 2

The Carmichael function $\lambda(40320)$ $=\lambda(2^7\cdot 3^2\cdot 5\cdot 7)$ $={\rm lcm}(\lambda(2^7), \lambda(3^2), \lambda(5), \lambda(7)) $ $={\rm lcm}(32, 6, 4, 6) =96$. So the value of $19^k\bmod 40320$ will cycle on a loop of length that divides $96$ and since $19$ is coprime to $40320$, we will have $19^{96}\equiv 1 \bmod 40320$.

Therefore $19^{10562} \equiv 19^{\large 10562 \bmod 96} \equiv 19^2\equiv 361 \bmod 40320.$ We can take e.g $k=0,1,2$ in $361+40320k$ to get three values as required.