Define a proper congruent number as an integer which represents the area of some right triangle $\triangle ABC$ whose legs are both strictly in $\mathbb{Q}/\mathbb{Z}$. For every proper congruent number there exist $a$, $b$, $c$, $d$, $e$, and $f$ such that
\begin{equation} \left(\dfrac{a}{b}\right)^2 + \left(\dfrac{c}{d}\right)^2 = \left(\dfrac{e}{f}\right)^2, \end{equation}
with $d\mid a $ and $b\mid c$. Is it true that $f = bd$ and $e^2 = (ad)^2+(bc)^2$? If so we see that each proper congruent number corresponds to some Pythagorean triple that has non square free components.
Suppose $n$ is a proper congruent number with $1=\gcd(a,b)=\gcd(c,b)=\gcd(e,f)$. Since $n$ is congruent it represents the area of the triangle $\triangle ABC$ which has area $n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d}) = \frac{1}{2}\frac{ac}{bd}$. Since $n$ is an integer it must follow that $bd\vert ac$. Since $n$ is proper we know that $b$ must divide $c$ and $d$ must divide $a$. Let $i$ denote the largest power of $b$ so that $b^{i}\vert c$ but $b^{i+1} \not \vert c$. Similarly, let $j$ denote the largest power of $d$ so that $d^{j}\vert a$ but $d^{j+1} \not \vert a$. Write $a =d^{j}q_2$ and $c =b^iq_1$ for some quotients $q_1$ and $q_2$, then by the hypothesis it follows that
\begin{equation} \label{Equation 1} \begin{split} (\frac{a}{b})^2+(\frac{c}{d})^2 & =(\frac{e}{f})^2\\ & \Downarrow\\ (\frac{d^{j}q_2}{b})^2+(\frac{b^iq_1}{d})^2 & =(\frac{e}{f})^2\\ & \Downarrow\\ \frac{d^{2j}q_2^2}{b^2}+\frac{b^{2i}q_1^2}{d^2} & =\frac{e^2}{f^2}\\ & \Downarrow\\ \frac{d^{2j+1}q_2^2+ b^{2i+1}q_1^2}{b^2d^2} & =\frac{e^2}{f^2}\\ \end{split}\tag{1} \end{equation}
Now suppose that the numerator and the denominator of the last equality in Eq.(1) are not in lowest terms. Then it must follow that $$b^2d^2\vert \left(d^{2j+1}q_2^2+ b^{2i+1}q_1^2\right)$$ but $bd\vert b^2d^2$ so then $$bd\vert b^2d^2\vert \left(d^{2j+1}q_2^2+ b^{2i+1}q_1^2\right).$$ Thus the product $bd$ divides the numerator. This implies $b$ and $d$ divide the numerator individually, but this cannot happen because $b$ divides the numerator if and only if $b\vert q_2^2$ (as $\gcd(b,d) = 1$ as $d$ is a divisor of $a$), but $a=d^jq_2$ and $1=\gcd(a,b)=\gcd(b,d^jq_2)$, which means $b$ cannot divides $q_2$. The same type of argument holds for the integer $d$. Thus the numerator and the denominator are in lowest terms, which implies that is $f^2 = b^2d^2 = (bd)^2$. Hence, $f=bd$ as desired.