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.

Question

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).

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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, $$[f(a)]^2 = \left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > (2\sqrt2)^2 = 8 \tag{5}$$

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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.

Answer 1

If $(a,b,c) \in \mathbb{N}^3$ is a Pythagorean triple, then without loss of generality, we may assume that $a=\left(m^2-n^2\right)d$, $b=2mnd$, and $c=\left(m^2+n^2\right)d$ for some positive integers $m,n,d$ with $m>n$, $\gcd(m,n)=1$, and $m\not\equiv n\pmod{2}$.

We now have $$t:=\frac{c}{a}+\frac{c}{b}=\frac{m^2+n^2}{m^2-n^2}+\frac{m^2+n^2}{2mn}=\frac{\left(m^2+n^2\right)\left(m^2-n^2+2mn\right)}{2mn\left(m^2-n^2\right)}\,.$$ Note that, since $t\in\mathbb{Q}$, we see that $t^2\in\mathbb{Z}$ iff $t\in\mathbb{Z}$. Now, $t\in\mathbb{Z}$ implies that $2mn$ divides $\left(m^2+n^2\right)\left(m^2-n^2+2mn\right)$. Since $\gcd(m,n)=1$ and $m\not\equiv n\pmod{2}$, we see that $\gcd\left(m^2+n^2,2mn\right)=1$. Therefore, $2mn$ must divide $m^2-n^2+2mn$, which means $2mn\mid m^2-n^2$. Again, we have $\gcd\left(2mn,m^2-n^2\right)=1$, which leads to a contradiction. Hence, $t\notin\mathbb{Z}$, so $t^2\notin\mathbb{Z}$.

Note that $t$ can get arbitrarily close to $2\sqrt{2}$. We can find $m$ and $n$ such that $\frac{m}{n}$ is arbitrarily close to $1+\sqrt{2}$. If $r:=\frac{m}{n}$, then $$t=\frac{\left(r^2+1\right)\left(r^2-1+2r\right)}{2r\left(r^2-1\right)}\,,$$ so as $r\to 1+\sqrt{2}$, $t\to 2\sqrt{2}$.

Answer 2

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).

Answer 3

First note that if $n$ were a square of rational number, then it would be square of integer. So if $(\frac{c}{a}+\frac{c}{b})^2$ were an integer, so would be $\frac{c}{a}+\frac{c}{b}$. By dividing by common factor, we can assume that triple $(a,b,c)$ is primitive, i.e. no two numbers share a prime factor.

Now $\frac{c}{a}+\frac{c}{b}=\frac{ca+cb}{ab}$. If this were an integer, we would have (in particular) $a\mid ca+cb$, so $a\mid cb$. But $a,b,c$ share no prime factor, so we would need to have $a=1$. But $1$ is not element of any Pythagorean triple.