Solution:
Let $(a, b, c)$ be a Pythagorean triple. View $a, b$ as lengths of the legs of a right angled triangle with hypotenuse of length $c$; let $\theta$ be the angle determined by the sides with lengths $a$ and $c$. Then
$$\begin{aligned} \left(\frac{c}{a}+\frac{c}{b}\right)^{2} & =\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)^{2}=\frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta}{(\sin \theta \cos \theta)^{2}} \\ & =4\left(\frac{1+\sin 2 \theta}{\sin ^{2} 2 \theta}\right)=\frac{4}{\sin ^{2} 2 \theta}+\frac{4}{\sin 2 \theta} \end{aligned}$$
Note that because $0<\theta<90^{\circ}$, we have $0<\sin 2 \theta \leq 1$, with equality only if $\theta=45^{\circ}$. But then $a=b$ and we obtain $\sqrt{2}=c / a$, contradicting $a, c$ both being integers. Thus, $0<\sin 2 \theta<1$ which gives $(c / a+c / b)^{2}>8$.

Defining $\theta$ as in Solution 1, we have $c / a+c / b=\sec \theta+\csc \theta$. By the AM-GM inequality, we have $(\sec \theta+\csc \theta) / 2 \geq \sqrt{\sec \theta \csc \theta}$. So
$$c / a+c / b \geq \frac{2}{\sqrt{\sin \theta \cos \theta}}=\frac{2 \sqrt{2}}{\sqrt{\sin 2 \theta}} \geq 2 \sqrt{2}$$
Since $a, b, c$ are integers, we have $c / a+c / b>2 \sqrt{2}$ which gives $(c / a+c / b)^{2}>8$.

By simplifying and using the AM-GM inequality,
$$\left(\frac{c}{a}+\frac{c}{b}\right)^{2}=c^{2}\left(\frac{a+b}{a b}\right)^{2}=\frac{\left(a^{2}+b^{2}\right)(a+b)^{2}}{a^{2} b^{2}} \geq \frac{2 \sqrt{a^{2} b^{2}}(2 \sqrt{a b})^{2}}{a^{2} b^{2}}=8$$
with equality only if $a=b$. By using the same argument as in Solution $1, a$ cannot equal $b$ and the inequality is strict.

$$\begin{aligned} \left(\frac{c}{a}+\frac{c}{b}\right)^{2} & =\frac{c^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}+\frac{2 c^{2}}{a b}=1+\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+1+\frac{2\left(a^{2}+b^{2}\right)}{a b} \\ & =2+\left(\frac{a}{b}-\frac{b}{a}\right)^{2}+2+\frac{2}{a b}\left((a-b)^{2}+2 a b\right) \\ & =4+\left(\frac{a}{b}-\frac{b}{a}\right)^{2}+\frac{2(a-b)^{2}}{a b}+4 \geq 8, \end{aligned}$$
with equality only if $a=b$, which (as argued previously) cannot occur.

Since $c / a+c / b$ is rational, $(c / a+c / b)^{2}$ can only be an integer if $c / a+c / b$ is an integer. Suppose $c / a+c / b=m$. We may assume that $\operatorname{gcd}(a, b)=1$. (If not, divide the common factor from $(a, b, c)$, leaving $m$ unchanged.)
Since $c(a+b)=m a b$ and $\operatorname{gcd}(a, a+b)=1$, $a$ must divide $c$, say $c=a k$. This gives $a^{2}+b^{2}=a^{2} k^{2}$ which implies $b^{2}=\left(k^{2}-1\right) a^{2}$. But then $a$ divides $b$ contradicting the fact that $\operatorname{gcd}(a, b)=1$. Therefore $(c / a+c / b)^{2}$ is not equal to any integer $n$.

We begin as in Solution 1, supposing that $c / a+c / b=m$ with $\operatorname{gcd}(a, b)=1$. Hence $a$ and $b$ are not both even. It is also the case that $a$ and $b$ are not both odd, for then $c^{2}=a^{2}+b^{2} \equiv 2(\bmod 4)$, and perfect squares are congruent to either 0 or 1 modulo 4. So one of $a, b$ is odd and the other is even. Therefore $c$ must be odd.
Now $c / a+c / b=m$ implies $c(a+b)=m a b$, which cannot be true because $c(a+b)$ is odd and $m a b$ is even.