<s>[INST] 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!
 [/INST] Because $2226$ divided by $7$, $5551$ divided by $7$, 
$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1})$, for odd $n$ we have $a^n+b^n=(a+b)(a^{n-1}-...+b^{n-1})$ and
$$2222^{5555}+5555^{2222}=$$
$$=(2226-4)^{5555}+4^{5555}+$$
$$+(5551+4)^{2222}-4^{2222}+$$
$$-\left(4^{5555}-4^{2222}\right)$$
and $$4^{5555}-4^{2222}=4^{2222}\left(4^{3333}-1\right),$$
where the last expression divided by $4^3-1$, which divided by $7$.
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