Suppose you have a rational point $\Big( \frac{a}{b}, \frac{c}{d} \Big)$ on the unit circle $x^2 + y^2 = 1$ such that both coordinates are in lowest terms. What is the quickest way to show that $b$ and $d$ must both be odd? (if possible, without resorting to the Pythagorean triple standard form construction).
I aim to assign a problem to my students which would require them to prove this fact, but we have not covered the proof for the standard form of Pythagorean triples.
On
$$(ad)^2+(bc)^2=(bd)^2\;.$$
First, $b$ and $d$ must have the same parity, because if, say, $b$ were even and $d$ were odd, then $a$ would be odd (since the fractions are in lowest terms), so the equation would say odd plus even is even.
If $b$ and $d$ are both even, the factors of $2$ on both sides can't match up. Say $b=r2^j$ and $d=s^k$, with $r,s$ odd and $j,k\gt0$. If $j\ne k$, say, without loss of generality $j\lt k$, then the left-hand side only has $2j$ factors of whereas the right-hand side has $2(j+k)$; whereas if $j=k$, the equation reads
$$ \left((as)^2+(rc)^2\right)2^{2j}=(rs)^22^{4j}\;. $$
As $a$ and $c$ are odd (since the fractions are in lowest terms), the left-most factor is the sum of two odd squares, and thus has residue $2$ modulo $4$ (as each odd square has residue $1$). Thus it contains exactly one factor of $2$, and $2j+1\lt4j$.
The condition is equivalent to $a^2d^2 + c^2 b^2 = b^2d^2$.
Suppose $b$ is even and $b = 2^k m$, with $k \geq 1$ and $m $ odd. Then $a$ is odd and $a^2d^2 + 2^{2k} m^2c^2 = 2^{2k} m^2d^2$, which implies that $2^k$ divides $d$. Reverse the roles of $b$ and $d$ to conclude that $b = 2^k m$ and $d = 2^k n$ with $k \geq 1$ and $m, n$ both odd. But then $(ma)^2 + (nc)^2 = m^2a^2 + n^2c^2 = 2^{2k} m^2n^2 \equiv 0 \pmod{4}$, which is impossible, since $ma$ and $nc$ are odd numbers and $(odd)^2 \equiv 1 \pmod{4}$.
Therefore $b$ is odd and, by symmetry, $d$ is odd as well.