Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Fermat's Little Theorem

Let $a$ be a positive integer and $p$ be a prime. Then $a^p\equiv a \pmod{p}$.
The cancellation rule tells us that we can divide by $a$ if $gcd (a,p) = 1$
$i.e., gcd(a,p) = 1$ $ \Rightarrow$ $a^{p-1}\equiv 1 \pmod{p}$

Proof:

Now, suppose that $gcd(a,p) = 1.$ We form the sequence
$ a, 2a, 3a,....(p-1)a$                                                                       $...(1)$
No two of its terms are congruent mod $p$, since
$i \cdot a \equiv k \cdot a \pmod{p}$ $ \Rightarrow$  $i\equiv k \pmod{p}$  $ \Rightarrow$ $ i = k$

Hence , each of the numbers in $(1)$ is congruent to exactly one of the numbers
$1, 2, 3, ...p-1.$
This implies that,  $ a^{p-1} \cdot1 \cdot2...(p-1) $ $\equiv 1 \cdot 2...(p-1) \pmod{p}$

 We may cancel with $(p-1)!$, since $(p-1)!$ and $p$ are coprime. 
Thus, $a^{p-1}\equiv 1 \pmod{p}$

 Note: The converse of Fermat's Little theorem is not valid
i.e., if for any integer $a^p\equiv a \pmod{p}$, it does not imply that $p$ is prime. 

 Example :

Find the remainder when $2^{32}$ is divided by $7$

 Solution : 

 Let $a = 2$  and $ p = 7$
Using Fermat's theorem, we have 
$2^{7-1}\equiv 1 \pmod{7}$
$2^6\equiv 1 \pmod{7}$
$(2^6)^5\equiv 1^5 \pmod{7}$
$2^{30}\equiv 1 \pmod{7}$
$2^{30} \cdot 2^2 \equiv 1 \cdot 2^2 \pmod{7}$
$2^{32}\equiv 4 \pmod{7}$

 Hence, the remainder when $2^{32}$ is divided by $7$ is $4$

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Equilateral and Equiangular Polygons

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