Three Regular Tessellations

Theorem
There exist exactly $3$ regular tessellations of the plane.

Proof
Let $m$ be the number of sides of each of the regular polygons that form the regular tessellation.

Let $n$ be the number of those regular polygons which meet at each vertex.

From Internal Angles of Regular Polygon, the internal angles of each polygon measure $\dfrac {\paren {m - 2} 180^\circ} m$.

The sum of the internal angles at a point is equal to $360^\circ$ by Sum of Angles between Straight Lines at Point form Four Right Angles

So:

But $m$ and $n$ are both greater than $2$.

So:
 * if $m = 3$, $n = 6$.
 * if $m = 4$, $n = 4$.
 * if $m = 5$, $n = \dfrac {10} 3$, which is not an integer.
 * if $m = 6$, $n = 3$.

Now suppose $m > 6$.

We have:

But there are no integers between $2$ and $3$, so $m \not > 6$.

There are $3$ possibilities in all.

Therefore all regular tessellations have been accounted for.

Also see

 * Five Platonic Solids