Given that $a,b,c,d,e,f$ are all distinct positive integers, prove that there are no solutions for the following knowing their fractions are non-integers and completely simplified: $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{d}\right)^2 = e^2 \tag{1}$$
I coulden't find a proof for this one but I managed to prove this for the case of $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{d}\right) ^2 = \left(\frac{e}{f}\right)^2 \tag{2}$$
by simplifying:
$$\frac{(ac)^2 + (bd)^2}{(cd)^2} = (\frac{e}{f})^2 $$
Since $(ac)^2 + (bd)^2$ must satisfy a Pythagorean triple, we know that each side length(exculding the hypotenuse) must be the product of two distinct integers, which can be proven through Euclid's formula :
$$ac = (2m)(n)$$ $$bd = (m -n)(m + n)$$
Q.E.D.
I also noted that $(1)$ can be written as:
$$(ac)^2 + (bd)^2 = (ecd)^2$$
This lead me to conjecture that there are no Pythagorean triples where the hypotenuse has 2 distinct integer factors where one of each is shared with its side lengths. Furthermore, I also conjecture the hypotenuse's remainder must be a prime such that those conditions can never be satisfied.
Can anyone prove or disprove this?
Suppose the equation $\left(\frac ab\right)^2+\left(\frac cd\right)^2 = e^2$ holds, where the fractions are in their simplest form, i.e. $\gcd(a,b) = \gcd(c,d) = 1$. Then $$\frac {a^2}{b^2} = \frac {e^2d^2 - c^2}{d^2}$$
Both sides are in their simplest form since $$\gcd(a^2,b^2) = 1 = \gcd(c^2,d^2) = \gcd(e^2d^2 - c^2, d^2)$$
so $a^2 = e^2d^2-c^2, b^2 = d^2$.
Given $b,d > 0$, $b=d$. Therefore no solutions exists, and this fact is unrelated to Pythagorean triples.