The solution to this equation is called Pythagorean triple because of its geometric meaning related to the Pythagorean theorem. All the solutions to $a^2 + b^2 = c^2$ can be expressed in the following way: $$\left\{\begin{align}a&=m^2 - n^2\\b&=2mn\\c&=m^2+n^2\end{align}\right.$$ when $m > n$ are two positive integers.

It is clear that if $(a, b, c)$ is a Pythagorean triple, so will be $(ka, kb, kc)$ where $k$ is any positive integer greater than $1$. Therefore, we are more interested in so called primitive Pythagorean triple where $a$, $b$, $c$ are co-prime. There are in total $16$ primitive Pythagorean triples where $c < 100$.

Primtive Pythagorean triples have many interesting properties. For example:

• Exactly one of $a$ and $b$ is divisible by $2$ (i.e. even).
• Exactly one of $a$ and $b$ is divisible by $3$.
• Exactly one of $a$ and $b$ is divisible by $4$.
• Exactly one of $a$ and $b$ is divisble by $5$.
• All prime factors of $c$ are in the form of $4k+1$. Hence, $c$ itself is of the form $4k+1$.
• The value of $\frac{(c-a)(c-b)}{2}$ must be a perfect square.
• At most one of $a$, $b$, $c$ is a square

• Fermat's Last Theorem: there is no integer solution to $x^n+y^n=z^n$ when $n > 2$.
• How to derive the solution