Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic) I tried using the following approach:
$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$
Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,
$$x=(2222^5+5555^2)k,\ k\in N$$
$$x=(1111^2)(32\cdot1111^3+25)k$$
The term in parentheses was found to be $54165190296027657$, which is divisible by $7$. Thus $7 | x$.
Another method (edit):
We can also write $$2222^{5555} = (7\times 317+3)^{5555} = 7p+3^{5555}$$
Similarly $$5555^{2222}=(793\times7+4)^{2222} = 7q+4^{2222}$$
So we have $$x=7(p+q)+(3^5+4^2)r$$
Here we have $3^5+4^2$ is divisible by $7$. So $x$ is also divisible by $7$.
But is there a simple way without multiplication of large numbers? Thank you!

We have
\begin{align}
2222^5 + 5555^2&=2222^5 +5555^5-5555^5+ 5555^2\\
&=2222^5 +5555^5-5555^2( 5555^3-1)\\
\end{align}
$2222^5 +5555^5$ is divisible by $2222+5555=7(1111)$ as $5$ is odd.
$$5555^3-1=(7\times793+4)^3-1=7p+4^3-1=7p+63$$
is divisible by $7$.
