1

What I did is, the product is also a natural number.

$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$

Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$

Hence $2x$ and $2y$ are natural numbers.

I can't proceed further. Is my way correct? Any help please.

Bill Dubuque
  • 272,048
Madhan Kumar
  • 826
  • 1
    Not only does $\sqrt{m^2-4}$ have to be rational. It must be an integer. And $\sqrt{m^2-4}=\sqrt{(m+2)(m-2)}$. $\gcd(m+2, m-2) =1,2,4$ so $\frac {m+2}{1,2,4}$ and $\frac {m-2}{1,2,4}$ must both be perfect squares but only $1,2,$ or $4$ apart. – fleablood Mar 22 '21 at 17:26
  • Either $x$ and $y$ is natural numbers or is of the form $\frac{p}{2}$. Suppose one of them is natural, then we are done. What about the other case? – Madhan Kumar Mar 22 '21 at 17:28

2 Answers2

1

If $x,y$ are rational $x = \frac pq$ and $y = \frac ab$ (assume in lowest terms) so

$\frac pq + \frac ba = \frac {pa + qb}{aq}\in \mathbb Z$

And $\frac ab +\frac qp = \frac {pa+qb}{pb}\in \mathbb Z$

So $q|pa+qb$ so $q|pa$ but $q$ and $p$ are relatively prime so $q|a$. Likewise $a|pa + qb$ so $a|qb$ so $a|q$ so $a= q$. And similarly $p = b$.

So $x = \frac 1y$.

So we must have $x + \frac 1y = x + x = 2x$ and $y + \frac 1x = 2y$ are natural numbers.

Whoo boy.

Okay. Case one $x = 1 \in \mathbb Z$ then $y = \frac 11 = 1$ and $x = y = 1$ are obviously solutions.

Case 2: $x \in \mathbb Z$ but $x > 1$. Then $y=\frac 1x \not \in \mathbb Z$. But $2y=\frac 2x\in \mathbb N$ so $x|2$ so $x = 2$ and $y= \frac 12$ is a solutions ($2+ \frac 1{\frac 12} = 4$ and $\frac 12 + \frac 12 = 1$).

Case 3: $x\not \in \mathbb Z$ then $2x \in \mathbb Z$ so $x = \frac m2$ for some odd $m$. But then $y =\frac 2m$ and $2y=\frac 4m$ is an integer so $m|4$. But $m$ is odd so $m =1$.

So those are the only three possibilities.

$x = y = 1; x = 2; y =\frac 12$ and $x = \frac 12$ and $y = 2$.

fleablood
  • 124,253
0

Hint If $\sqrt{m^2-4}=\frac{p}{q}$ is rational with $\mbox{gcd}(p,q)=1$, then $\frac{p^2}{q^2} \in \mathbb Z$. Use $\mbox{gcd}(p,q)=1$ to conclude that $q=1$.

This gives that $\sqrt{m^2-4}=n$ for some $n \in \mathbb N$ and hence $$ 4=m^2-n^2=(m-n)(m+n) $$ Look at all possible factorizations to conclude that $m=2$.

To complete the problem, observe that $2x=n, 2y=k$ and $xy=1$ gives $$ kn=4 $$ There are only 3 possibilities (or two by symmetry), just check them.

N. S.
  • 132,525