Prove that $2222^{5555}+5555^{2222}=3333^{5555}+4444^{2222} \pmod 7$

Question

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.

Answer 1

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 $2222^{5555} + 5555^{2222} = 3333^{5555} + 4444^{2222}$.

Answer 2

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. :\

Answer 3

$$2222=11\cdot202=22\cdot 101=101\pmod 7\\5555=11\cdot505=55\cdot 101=-101\pmod 7\\3333=11\cdot 303=33\cdot 101=(-2)\cdot 101\pmod 7\\4444=11\cdot404=44\cdot 101=2\cdot101\pmod 7$$

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

$$\begin{align*}2222&=&3\\5555&=-3=&4\\3333&=-2\cdot 3=-6=&1\\4444&=2\cdot 6=&-1\end{align*}$$

Answer 4

See cycles.

1. $2222^6 \equiv 1\pmod{7}$ and $5555\equiv 5\pmod{6}$ then $2222^{5555}\equiv 2222^5\pmod{7}\equiv 5\pmod{7}$
2. $5555^3 \equiv 1\pmod{7}$ and $2222\equiv 2\pmod{3}$ then $5555^{2222}\equiv 5555^2\pmod{7}\equiv 2\pmod{7}$

left hand $= 5 + 2 \equiv 0\pmod{7}$

1. $3333^1 \equiv 1\pmod{7}$ and $5555\equiv 0\pmod{1}$ then $3333^{5555}\equiv 3333^1\pmod{7}\equiv 1\pmod{7}$

2. $4444^1 \equiv 1\pmod{7}$ and $2222\equiv 0\pmod{1}$ then $4444^{2222}\equiv 4444^1\pmod{7}\equiv 1\pmod{7}$

right hand $= 1 + 1 \equiv 2\pmod{7}$

$2222^{5555}+5555^{2222}=3333^{5555}+4444^{2222} \pmod 7$ is incorrect.

You can check with bc program:

bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'. 
2222^5555 % 7
5
5555^2222 % 7
2
3333^5555 % 7
1
4444^2222 % 7
1