# Rules for algebra equations involving modulo operations

While working on a menial task in front of a clock today I was distracting myself by proving that all three hands only align twice a day. That lead me to wonder how one would deal with more complex problems involving modulo arithmetic. I know several rules for reducing equations involving all sorts of operators from simple addation up through very complex triple integrals and the like. But, I never learned any rules for manipulating the mudulo operator.

What are valid oberations that can be used to reduce eautaions involving multiple modulo operators?

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Are you familiar with rings, and fields, and that the integers mod $\rm\:m\:$ for a ring, and a field if $\rm\:m\:$ is prime? –  Math Gems Mar 30 '13 at 21:24
Gems, for the sake of argument, lets say I'm not. I actually wandered into set theory a little when I was learning Haskell, but I really only know enough to be dangerous. –  John F. Miller Mar 31 '13 at 0:33

Here are some examples I can think of.

Let $m$ be any natural number, and let $a,b,c,d$ be any integers. Then:

• $\equiv$ modulo $m$ is an equivalence relation. That is,
• $a\equiv a\bmod m$.
• If $a\equiv b\bmod m$, then $b\equiv a\bmod m$.
• If $a\equiv b\bmod m$ and $b\equiv c\bmod m$, then $a\equiv c\bmod m$.
• Addition and multiplication are well-defined modulo $m$. That is,
• If $a\equiv b\bmod m$ and $c\equiv d\bmod m$, then $a+c\equiv b+d\bmod m$, and $ac\equiv bd\bmod m$.
• If $ac\equiv bc\bmod mc$, then $a\equiv b\bmod m$.
• The congruence $ax\equiv b\bmod m$ has solutions (i.e., integers $x$ making the statement true) if and only if $\gcd(a,m)$ divides $b$.

You also have

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