Prove that $2222^{5555}+5555^{2222}=3333^{5555}+4444^{2222} \pmod 7$ I am utterly new to modular arithmetic and I am having trouble with this proof.
$$2222^{5555}+5555^{2222}=3333^{5555}+4444^{2222} \pmod 7$$
It's because $2+5=3+4=7$, but it's not so clear for me with the presence of powers.
Maybe some explanation would help.
EDITED Some serious typo
EDIT
Since some arguments against it appear here is :
WolframAlpha
EDIT Above is incorrect. I appreciate proofs that it is wrong. Sorry for others.

First recall that as $7$ is prime, then $x^6 = 1 \pmod{7}$. Now, we have
$$ 2222 = \begin{cases} 2 \pmod{6} \\ 3 \pmod{7} \end{cases}, \quad 3333 = 1 \pmod{7}$$
$$4444 = -1 \pmod{7}, \quad 5555 = \begin{cases} 5 \pmod{6} \\ 4 \pmod{7} \end{cases}$$
Then we can reduce each side of the equation to 
$$ 3^5 + 4^2 = 1^5 + (-1)^2 \pmod{7}$$
Then the LHS is $0$ but the RHS is $2$, so the statement is false.
EDIT: For reference, I'm testing the conjecture $2222^{5555} + 5555^{2222} = 3333^{5555} + 4444^{2222}$.
