<s>[INST] Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer. 
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there are no integer $n$ and Pythagorean
  triple $(a, b, c)$ satisfying $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n$$ 

The foregoing question is from the 2005 Canada National Olympiad.
I need some help on part (b).

Part (a)
Examine the behaviour of the following function [that embodies the constraint $c^2=a^2+b^2$, $x$ can play the role of either $a$ or $b$] on $(0,c)$:
$$f(x) = \dfrac{c}{x} + \dfrac{c}{\sqrt{c^2-x^2}} \tag{1}$$
Now find the derivative: $$f'(x)=cx\left[\dfrac{1}{(c^2-x^2)^{3/2}}-\dfrac{1}{x^3}\right] \tag{2}$$
To find local extrema: 
$$\begin{align}
f'(x)=0 &\implies x^3=(c^2-x^2)^{3/2} \\
&\implies x^6=(c^2-x^2)^3 &(\text{squaring}) \\
&\implies u^3=(k-u)^3 &(u=x^2,k=c^2) \\
&\implies 2u^3-3ku^2+3k^2u-k^3=0 \\
&\implies (2u-k)(u^2+-ku+k^2)=0 \tag{3}\\
\end{align}$$
So, solutions are: $$u=\dfrac{k}{2};u=\dfrac{k\pm\sqrt{k^2-4k^2}}{2}\notin\mathbb{R}$$
So take the only real solution as $$u=\dfrac{k}{2} \implies x^2=\dfrac{c^2}{2} \implies x=\dfrac{c}{\sqrt2}\quad(\text{since }x\in(0,c))$$
This is a local minima because $f$ is continuous on $(0,c)$ and $$\lim_{x\to0^+}{f(x)}=\lim_{x\to c^-}{f(x)}=+\infty$$
The minimum value is $$f\left(\dfrac{c}{\sqrt2}\right)=\sqrt2+\sqrt2=2\sqrt2 \tag{4}$$
This value is not achievable, because $\sqrt{2}$ is not a rational number. Hence,
$$[f(a)]^2 = \left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > (2\sqrt2)^2 = 8 \tag{5}$$

Part (b)
$$\begin{align}
\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n \implies c^2(a^2+b^2)=a^2b^2\cdot n \tag{6}
\end{align}$$
in a perhaps friendlier form.
I am not sure how to use this further.
 [/INST] For now, I prove you part (a) with a different approach. Dividing by $c^2=a^2+b^2$, taking the square root, and multiplying by $-1/2$, the inequality can be rewritten as
$$
\frac{2}{\frac{1}{a}+\frac{1}{b}}<\sqrt{\frac{a^2+b^2}{2}},
$$
which is simply HM-QM. Equality cannot hold because it cannot be the case that $a=b$ (indeed $2a^2$ cannot be a square).
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