# Square of Pythagorean Prime is Hypotenuse of Pythagorean Triangle

## Theorem

Let $p$ be a Pythagorean prime.

Then $p^2$ is the hypotenuse of a Pythagorean triangle.

## Proof

By Fermat's Two Squares Theorem, a Pythagorean prime, $p$ can be expressed in the form:

$p = m^2 + n^2$

where $m$ and $n$ are (strictly) positive integers.

$(1): \quad m \ne n$, otherwise $p$ would be of the form $2 m^2$ and so even and therefore not a prime.
$(2): \quad m \perp n$, otherwise $\exists c \in \Z: p = c^2 \left({p^2 + q^2}\right)$ for some $p, q in \Z$ and therefore not a prime.
$(3): \quad m$ and $n$ are of opposite parity, otherwise $p = m^2 + n^2$ is even and therefore not a prime.

Without loss of generality, let $m > n$.

From Solutions of Pythagorean Equation, the triple $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ forms a primitive Pythagorean triple such that $m^2 + n^2$ forms the hypotenuse of a primitive Pythagorean triangle.

Thus we have that $p^2 = r^2 + s^2$ for some (strictly) positive integers $r$ and $s$.

Similarly, we have that $r \ne s$, otherwise $p^2$ would be of the form $2 r^2$ and so not a square number.

Without loss of generality, let $r > s$.

From Solutions of Pythagorean Equation, the triple $\left({2 r s, r^2 - s^2, r^2 + s^2}\right)$ forms a Pythagorean triple such that $p^2 = r^2 + s^2$ forms the hypotenuse of a Pythagorean triangle.

$\blacksquare$