Recall:

• a mod m means the remainder upon dividing by m; this partitions the integers into equivalence classes.
• a ≡ b (mod n) means… m | a-b.
• In +,⋅ mod m, we saw we can replace any number with a congruent number. Not so for exponentiation.

Division mod m

Consider the series 0, 1⋅a, 2⋅a, 3⋅a, 4⋅a, … mod m.

• Try a=5, m=8:
  0, 5, 2, 7, 4, 1, 6, 3, 0, 5, 2, 7, …
  We see we hit a cycle (of length 8). What is 5^-1 mod 8? That is, the multiplicative inverse of 5, mod 8? Looking at the sequence above, we see that 5⋅5≡1, so 5 is its own inverse: 5^-1 ≡ 5 (mod 8).
• Try a=6, m=8:
  0, 6, 4, 2, 0, 6, …
  Uh-oh, we hit a cycle (of length four). What is 6^-1 mod 8? It doesn't exist!
  (This makes sense when you think about it: 6 times any number will be even, so it will never by 1 plus a multiple of 8.

Th'm: If ac≡bc (mod m), and gcd(c,m)=1, then a≡b (mod m).

What this means:
"We can divide by c, mod m." For instance, let m=8 and c=3. Knowing that 3x ≡ 15 (mod 8), we can divide both sides by 3 and conclude that x ≡ 5. (Indeed, this turns out to be the the only x satisfying the constraint, mod 8).
However, consider 4x &equiv 12 (mod 8). We can not conclude that x≡3. In fact, only x≡1 satisfies the constraint 4x≡12 (mod 8).

Applications: parity check; hash: ssn mod 10⁴.