$a/b + b/a$ is an integer if and only if $a = b$

Question

So for an iff, I know you must prove it both ways. I have proven the converse by the idea that

$\frac{a}{b} + \frac{b}{a} = \frac{a}{a} + \frac{a}{a} = 2 $ which is an integer.

But I am struggling with the direct proof that if $\frac{a}{b} + \frac{b}{a}$ is an integer then $a = b$.

I have gotten to $ab \mid a^2 + b^2$ but I end up trying to justify why $\frac{b^2}{a}$ is not an integer which I cannot even prove intuitively.

What are a and b ? real numbers , integers, natural number ?? — Murtuza Vadharia, Apr 21 '16 at 09:54
See also https://math.stackexchange.com/questions/1468189/show-that-if-a-neq-b-and-a-and-b-are-positive-then-fracab-fracba and https://math.stackexchange.com/questions/1626241/prove-a-bb-a-for-a-and-b-natural-is-only-natural-for-a-b — Martin Sleziak, Aug 11 '17 at 11:27

score 3 · Answer 1 · answered Apr 21 '16 at 09:57

3

HINT:

Let $\dfrac ab+\dfrac ba=k\iff\left(\dfrac ab\right)^2-k\dfrac ab+1=0$

$\implies\dfrac ab=\dfrac{k\pm\sqrt{k^2-4}}2$

We need $k^2-4$ to be perfect square $=r^2$(say)

$(k+r)(k-r)=4$

answered Apr 21 '16 at 09:57

lab bhattacharjee

1

Observe that $k\pm r$ have the same parity – lab bhattacharjee Apr 21 '16 at 10:03
Okay I got to the last step. I can't quite see why this proves the idea that a=b. – Hilly Doctorson Apr 21 '16 at 10:12
@Hilly, what are the possible values of $k$ ? – lab bhattacharjee Apr 21 '16 at 10:36

Roman83 · Answer 2 · 2016-04-21T10:33:00.810

2

$\gcd(a,b)=1$ (Otherwise, the numerator and denominator can be reduced)

Let $$\frac ab+ \frac ba =m, m \in \mathbb Z$$ Then $$a^2+b^2=abm$$ Then $a|b \Rightarrow a=1$. Similarly,$b=1$

edited Apr 21 '16 at 10:33

answered Apr 21 '16 at 09:55

Roman83

So we know that a|(a^2+b^2) and b|(a^2+b^2) and that gcd(a,b)=1. How did you get to a|b? And if so, does that prove the direct? – Hilly Doctorson Apr 21 '16 at 10:09
$b^2=a(bm-a) \Rightarrow a|b^2$, but $\gcd(a,b)=1$, then $a|b$ and $a=1$ – Roman83 Apr 21 '16 at 10:26
$\gcd(a,b)=1$ (Otherwise, the numerator and denominator can be reduced) – Roman83 Apr 21 '16 at 10:33

2 Answers2