The Pythagorean Triple Two Ways!

Let the Pythagorean triple $(m,n,o)$ be $m<n<o$ . Then we have to prove if there is solution for

$\begin{aligned} \frac 1{o^2} + \frac 1{n^2} &= \frac 1{m^2} \\ \frac {o^2+n^2}{o^2n^2} & = \frac 1{m^2} & \small \blue{\text{Note that }m^2+n^2=o^2} \\ \frac {m^2+2n^2}{m^2n^2+n^4} & = \frac 1{m^2} \\ m^4 + 2m^2n^2 & = m^2n^2+n^4 & \small \blue{\text{Divide both sides by }m^2n^2} \\ \frac {m^2}{n^2} + 2 & = 1 + \frac {n^2}{m^2} \\ \frac {m^2}{n^2} + 1 - \frac {n^2}{m^2} & = 0 & \small \blue{\text{Multiply both sides by }\frac {m^2}{n^2}} \\ \left(\frac {m^2}{n^2}\right)^2 + \frac {m^2}{n^2} - 1 & = 0 \\ \frac {m^2}{n^2} & = \frac {\sqrt 5-1}2 & \small \blue{\text{Since }\frac {m^2}{n^2}>0} \end{aligned}$

There is no rational solution for $\dfrac mn$ , hence the required natural number triple $(m,n,o)$ does not exist.

Since $m:n:o=\dfrac{1}{a}:\dfrac{1}{b}:\dfrac{1}{c}$ , therefore we have to have $(ab) ^2+(ac) ^2=(bc)^2$ . So $ab=m^2-n^2, bc =2mn, ca =m^2+n^2$ for some relative prime integers $m$ and $n$ . Then $b(c+a)=2m^2, b(c-a)=2n^2\implies b^2(c^2-a^2)=4m^2n^2\implies b^4=4m^2n^2\implies b^2=2mn$ . Since $(a, b, c)$ is a Pythagorean triplet, therefore $a=p^2-q^2, b=2pq, c=p^2+q^2$ for some integers $p, q$ . This implies $mn=2p^2q^2$ . So one of $m, n$ must be an odd square multiple of $2$ , and the other must be an odd square. Let $m=2(2r+1)^2, n=(2s+1)^2$ . Then $b=2(2r+1)(2s+1), a=\dfrac{4(2r+1)^4-(2s+1)^4}{2(2r+1)(2s+1)}, c=\dfrac{4(2r+1)^4+(2s+1)^4}{2(2r+1)(2s+1)}$ . But then neither $a$ , nor $c$ is an integer, since the numerator of each of them is odd and the denominator is even. Hence there is no such triplet $(a, b, c)$ satisfying the conditions of the problem. That is, the number of such triplets is $\boxed 0$ .

For any triple (a,b,c), say the ratio is $x:y:z$ , where they are all relatively prime. They MUST all be relatively prime because if 2 of them have a common factor greater than 1, then the other will also have that factor by pythagorean theorem. Every pythagorean triple must have this 'primitive' triple, where all sides are relatively prime.

Now, if a triple had terms in a ratio of $\frac {1}{x}$ : $\frac {1}{y}$ : $\frac {1}{z}$ , it will be in a ratio of $yz:xz:xy$ . Obviously, this is not primitive, since as discussed above, if 2 have a common prime factor, the third does as well. However, while each pair does have a common factor, all 3 should not, as $x, y, z$ are relatively prime, and thus if $yz$ and $xz$ have z in common, $xy$ should NOT have z in common. Therefore, this cannot be a pythagorean triple, and thus the answer is 0 .