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 $(c / a+c / b)^{2}>8$.

b) Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a, b, c)$ satisfying $(c / a+c / b)^{2}=n$.

## a) Solution 1

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 $0 < \sin 2\theta \leq 1$, we have $\frac{4}{\sin ^{2} 2 \theta}+\frac{4}{\sin 2 \theta} > 8$. Therefore, $\left(\frac{c}{a}+\frac{c}{b}\right)^{2} > 8$.