Let $\tuple {A, B, C}$ be a Pythagorean Triple: :$A^2 + B^2 = C^2$ By the Pythagorean theorem, this equation describes the sides of a right triangle: :400px By the definitions of sine and cosine: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \frac A C }} {{eqn | l = \cos \theta       | r = \frac B C }} {{end-eqn}} That is, {{begin-eqn}} {{eqn | l = A       | r = C \sin \theta }} {{eqn | l = B       | r = C \cos \theta }} {{end-eqn}} Next, we invoke Equiangular Triangles are Similar and Proportion is Equivalence Relation to write: {{begin-eqn}} {{eqn | l = A       | o = \propto       | r = B \propto C }} {{eqn | ll= \leadsto       | l = C \sin \theta       | o = \propto       | r = C \cos \theta \propto C       | c = from above }} {{eqn | ll= \leadsto       | l = \sin \theta       | o = \propto       | r = \cos \theta \propto 1       | c = dividing throughout by $C$ }} {{end-eqn}} We construct the following restriction of $\tan \dfrac \theta 2$. From Shape of Tangent Function, $\tan \dfrac \theta 2: \openint 0 \pi \leftrightarrow \openint 0 {+\infty}$ is a bijection. Restrict $\tan \dfrac \theta 2$ on this interval so that its image is the set of strictly positive rational numbers: :$\tan \dfrac \theta 2: \tan^{-1} \sqbrk {\Q_{>0} } \cap \openint 0 \pi \leftrightarrow \Q_{>0}$ Then $\displaystyle \tan \frac \theta 2 = \frac p q$ for any $\dfrac p q \in \Q_{>0}$, where $\dfrac p q$ is the canonical form of a rational number. From the Double Angle Formulas: {{begin-eqn}} {{eqn | l = \sin \theta       | r = 2 \sin \frac \theta 2 \cos \frac \theta 2 }} {{eqn | l = \cos \theta       | r = \cos^2 \frac \theta 2 - \sin^2 \frac \theta 2 }} {{eqn | l = 2 \sin \frac \theta 2 \cos \frac \theta 2       | o = \propto       | r = \cos^2 \frac \theta 2 - \sin^2 \frac \theta 2 \propto 1       | c = substituting into the above proportion }} {{eqn | ll= \leadsto       | l = 2\tan \frac \theta 2       | o = \propto       | r = 1 - \tan^2 \frac \theta 2 \propto \sec^2 \frac \theta 2       | c = dividing throughout by $\cos^2 \dfrac \theta 2$ }} {{eqn | ll= \leadsto       | l = 2 \tan \frac \theta 2       | o = \propto       | r = 1 - \tan^2 \frac \theta 2 \propto 1 + \tan^2 \frac \theta 2       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = 2 \frac p q       | o = \propto       | r = 1 - \paren {\frac p q}^2 \propto 1 + \paren {\frac p q}^2 }} {{eqn | ll= \leadsto       | l = 2 \frac p q       | o = \propto       | r = 1 - \frac {p^2} {q^2} \propto 1 + \frac {p^2} {q^2} }} {{eqn | ll= \leadsto       | l = 2 p q       | o = \propto       | r = q^2 - p^2 \propto q^2 + p^2       | c = multiplying throughout by $q^2$ }} {{end-eqn}} Thus these proportions describe the sides of a right triangle: By the Pythagorean theorem, $\tuple {2 p q, q^2 - p^2, q^2 + p^2}$ is a Pythagorean triple. That $p, q \in \Z_{>0}$ follows from $\dfrac p q \in \Q_{>0}$. That $p \perp q$ follows from the assumption that $\dfrac p q$ was written in canonical form. That ''every'' triple is of this form follows from the bijectivity of the tangent function as restricted above. Recall $q^2 - p^2$ describes the side of a triangle, and so is positive. Then $q^2 - p^2 > 0$ and therefore $q > p$. It remains to be proven that: :this triple is primitive :$p$ and $q$ are of opposite parity. {{finish}}