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.
 A: First, you can change it to $3^{5555} + 4^{2222} \equiv 1^{5555} + 6^{2222} \pmod{7}$.
Then, notice that $3^6 \equiv 1 \pmod{7}$, $4^3 \equiv 1 \pmod{7}$, and $6^2 \equiv 1 \pmod{7}$. You can shrink the exponents to $3^5 + 4^2 \equiv 1 + 1 \pmod{7}$
From there, it's just computation:
$$3^5 + 4^2 \equiv 1 + 6^0 \pmod{7}$$
$$243 + 16 \equiv 2 \pmod{7}$$
$$259 \equiv 2 \pmod{7}$$
$$0 \not\equiv 2 \pmod{7}$$
So it's false after all (I misread the exponents at first, sorry).
Since you're new to modular arithmetic, this may not be clear, but for all $x \ne 0$, $x^6 \equiv 1 \pmod{7}$. See Euler's totient theorem. 
EDIT: Wow I'm sloppy today. Sorry. :\
A: $$2222=11\cdot202=22\cdot 101=101\pmod 7\\5555=11\cdot505=55\cdot 101=-101\pmod 7\\3333=11\cdot 303=33\cdot 101=(-2)\cdot 101\pmod 7\\4444=11\cdot404=44\cdot 101=2\cdot101\pmod 7$$
Well, now add modulo $\,7\,$ both sides of your equation...:)
Added: By the way, note that $\,101=3\pmod 7\,$ , so again doing arithmetic modulo $\,7\,$ we indeed get
$$\begin{align*}2222&=&3\\5555&=-3=&4\\3333&=-2\cdot 3=-6=&1\\4444&=2\cdot 6=&-1\end{align*}$$
A: 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 $2222^{5555} + 5555^{2222} = 3333^{5555} + 4444^{2222}$.
