Rational And Its Sum

As suggested by Calvin. Here's the rational root theorem (RRT) approach:

Suppose there exists a positive integer $n$ that satisfy this condition, then $r + \dfrac1r = n \Leftrightarrow r^2 + 1 = nr \Leftrightarrow r^2 - nr + 1 = 0$ .

Since we just want to find the rational roots of this equation, then RRT is sufficient. All the possible rational roots ( $r$ ) of the quadratic equation $r^2 - nr + 1 = 0$ are

$\pm \dfrac{\text{all the positive factors of the constant term}}{\text{all the positive factors of the leading coefficient}} \; = \; \pm \dfrac{\text{all the positive factors of 1}}{\text{all the positive factors of 1}} \; = \; \pm 1.$

So we just need to check whether $r = \pm 1$ yields a positive integer $n$ . The original question already shown that $r=1$ satisfy this condition and we're looking for other solutions.

We just need to check for $r = -1$ only, but upon substitution, we can see that $r + \dfrac1r$ is not a positive integer, so there's NO other solution.

Let $r=\frac{p}{q}$ where $gcd(p,q)=1$ and $q \neq 1$

$\frac{p}{q} + \frac{q}{p}=I$ whre $I \in \mathbb N$

$p^2 + q^2=pqI$

$\frac{p^2}{q} + q=pI$

$\frac{p^2}{q}=pI-q$

As we previously assumed that $p,q$ are co-prime so $q \nmid p^2$ .L.H.S. is not an integer but R.H.S. is an integer.This contradicts that no rational exists.

Suppose $r + \dfrac{1}{r} = n$ for some $n \in \mathbb{N}$ . (Note that by the AM-GM inequality we know that $n \ge 2$ .)

Then $r^{2} - nr + 1 = 0 \Longrightarrow r = \dfrac{n \pm \sqrt{n^{2} - 4}}{2}$ .

For $r$ to be rational we will require that $n^{2} - 4$ be a perfect square, i.e., that $n^{2} - 4 = m^{2} \Longrightarrow n^{2} - m^{2} = 4$ for some $m \in \mathbb{N \cup \{0\}}$ . But this is only the case for $(n,m) = (2,0)$ , which yields the unique solution $r = 1$ , and thus there is $\boxed{No}$ other such positive rational number $r$ .

(Note that $n^{2} - m^{2} = (n + m)(n - m) = 4$ with $n \gt m \ge 0$ implies that either

• (i) $n + m = 4, n - m = 1 \Longrightarrow 2n = 5$ , i.e., $n \notin \mathbb{N}$ , or

• (ii) $n + m = 2, n - m = 2 \Longrightarrow n = 2, m = 0$ .)