TITLE: Why does $6^x ≡ 2^{10-x} \pmod{11}$ when $0≤x≤10$?
QUESTION [5 upvotes]: I was messing around with my calculator earlier today.  I graphed the function $6^x \pmod{11}$, and I noticed a pattern, and I "discovered" the following:
$$6^x ≡ 2^{10-x} \pmod{11}$$
This works whenever $x$ is an integer between $0$ and $10$, inclusive.  Likewise, these also seem to work:
$$4^x ≡ 3^{10-x} \pmod{11}$$
$$5^x ≡ 9^{10-x} \pmod{11}$$
$$7^x ≡ 8^{10-x} \pmod{11}$$
$$10^x ≡ 1^{10-x} \pmod{11}$$
I have two main questions:

What causes the pairs $(1,10)$, $(2,6)$, $(3,4)$, $(5,9)$, and $(7,8)$?
Why do relationships like this exist in the first place?

REPLY [3 votes]: First note that, for any integer $x$,
$$6^x\equiv 2^{10-x}\bmod 11 \;\iff\; 12^x=6^x\cdot 2^x\equiv 2^{10}\bmod 11.$$
Then note that $12\equiv 1\bmod 11$, so that $12^x\equiv 1\bmod 11$ for any $x$, and also that $2^{10}\equiv 1\bmod 11$, which one can compute directly:
$$2^{10}=1024=(11\cdot 93)+1\equiv 1\bmod 11$$
or just appeal to Fermat's little theorem.
We can generalize your observation as follows: for any positive integer $n$, and for any integers $a$ and $b$ such that $ab\equiv 1\bmod n$, we have
$$a^x\equiv b^{\varphi(n)-x}\bmod n$$
for all $0\leq x\leq \varphi(n)$ (and indeed, for all integers $x$, because we can make sense of the multiplicative inverse of $a$ and $b$ modulo $n$).