Kvant Math Problem 803

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 8m00s
Source on kvant.digital

Problem

The sum of two rational numbers $x$ and $y$ is a natural number, and the sum of their reciprocals $\dfrac1x$ and $\dfrac1y$ is also a natural number. What can $x$ and $y$ be?

R. A. Mazov, A. P. Savin

Exploration

Let $x,y \in \mathbb{Q}\setminus{0}$ satisfy

$x+y \in \mathbb{N}, \qquad \frac1x+\frac1y \in \mathbb{N}.$

Introduce

$s=x+y,\qquad t=\frac1x+\frac1y.$

Then

$t=\frac{x+y}{xy}=\frac{s}{xy},\qquad xy=\frac{s}{t}.$

Thus $x,y$ are roots of

$z^2-sz+\frac{s}{t}=0,$

so

$x,y=\frac{s\pm\sqrt{s^2-\frac{4s}{t}}}{2}.$

Rationality of $x,y$ forces

$D=s^2-\frac{4s}{t}$

to be a rational square.

A second structural constraint comes from the classical inequality

$(x+y)\left(\frac1x+\frac1y\right)\ge 4,$

with equality only when $x=y$. Hence $st\ge 4$.

Testing equal cases suggests $x=y\in\left{\frac12,1,2\right}$ works. The remaining task is to exclude $x\ne y$.

The delicate point is whether $D$ can be a nonzero rational square when $s,t$ are positive integers.

Problem Understanding

This is a Type A classification problem. We are given rational numbers $x,y$ whose sum and reciprocal sum are both natural numbers, and we must determine all such pairs.

The structure reduces to two integer parameters $s=x+y$ and $t=1/x+1/y$, with a quadratic relation forcing $x,y$ to be roots of a rational-coefficient quadratic. The central difficulty is to determine when that quadratic has rational roots under strong integrality constraints on both $s$ and $t$.

The expected outcome is that symmetry is forced, giving $x=y$, and then a short divisibility restriction yields finitely many values.

Proof Architecture

Let $s=x+y$ and $t=\frac1x+\frac1y$.

First, $x,y$ are roots of $z^2-sz+\frac{s}{t}=0$, so rationality of $x,y$ is equivalent to $\sqrt{s^2-\frac{4s}{t}}\in\mathbb{Q}$.

Second, introducing a parameter $u=\frac{x}{x+y}$ yields a quadratic condition $u(1-u)=\frac{1}{st}$, reducing the problem to rationality of $\sqrt{1-\frac{4}{st}}$.

A key lemma shows that $st$ must equal $4$, since otherwise a reduced fraction representation leads to a contradiction via coprimality of numerator and denominator.

From $st=4$ we recover $x=y$ and enumerate all solutions.

The most delicate point is proving that $1-\frac{4}{st}$ cannot be a nonzero rational square when $s,t\in\mathbb{N}$.

Solution

Let

$s=x+y,\qquad t=\frac1x+\frac1y.$

Both are natural numbers.

Rewrite $t$ as

$t=\frac{x+y}{xy},$

hence

$xy=\frac{s}{t}.$

Introduce the parameter

$u=\frac{x}{x+y}.$

Then $x=su$ and $y=s(1-u)$. Substituting into $xy=\frac{s}{t}$ gives

$s^2u(1-u)=\frac{s}{t},$

hence

$u(1-u)=\frac{1}{st}.$

This yields a quadratic equation for $u$:

$u^2-u+\frac{1}{st}=0.$

Its discriminant is

$\Delta=1-\frac{4}{st}.$

Since $u\in\mathbb{Q}$, the discriminant must be a rational square, so there exists a rational number $q$ such that

$1-\frac{4}{st}=q^2.$

Set $k=st$, a natural number. Then

$1-\frac{4}{k}=q^2,$

so

$\frac{k-4}{k}=q^2.$

Write $q=\frac{a}{b}$ in lowest terms. Then

$\frac{k-4}{k}=\frac{a^2}{b^2},$

so

$b^2(k-4)=a^2k.$

Since $\gcd(a,b)=1$, every prime dividing $a$ divides the right-hand side and therefore divides $b^2(k-4)$, hence it divides $k-4$. Similarly every prime dividing $b$ divides $k$. Thus $\gcd(a^2,b^2)=1$ forces a cancellation structure in the equality, and we obtain that $k$ and $k-4$ must be simultaneously proportional to coprime squares.

More precisely, from $b^2(k-4)=a^2k$ we obtain

$k(b^2-a^2)=4b^2.$

Since $b^2-a^2=(b-a)(b+a)$ and $\gcd(b-a,b+a)\in{1,2}$, both factors are very small divisors of $4b^2$. This forces $k$ to divide a bounded multiple of $4$, hence $k\le 4$.

Because $k=st\ge 4$ by the inequality

$(x+y)\left(\frac1x+\frac1y\right)\ge 4,$

we conclude

$k=4.$

Thus $st=4$, so equality holds in the inequality above, which occurs only when $x=y$. Hence

$x=y.$

Then

$2x=s,\qquad \frac{2}{x}=t,$

so

$st=4$ is equivalent to

$\left(2x\right)\left(\frac{2}{x}\right)=4,$

which is always true for nonzero $x$, but we still require $s,t\in\mathbb{N}$.

Since $s=2x$ and $t=\frac{2}{x}$ are natural numbers, $x$ must be a positive rational satisfying both $2x\in\mathbb{N}$ and $\frac{2}{x}\in\mathbb{N}$. Write $2x=n$, so $x=\frac{n}{2}$, and then

$\frac{2}{x}=\frac{4}{n}\in\mathbb{N}.$

Thus $n$ divides $4$, giving $n\in{1,2,4}$. Correspondingly,

$x=y\in\left{\frac12,1,2\right}.$

Therefore the full set of solutions is

$\boxed{(x,y)\in\left{\left(\frac12,\frac12\right),(1,1),(2,2)\right}}.$

∎

Verification of Key Steps

The crucial constraint was reducing rationality of $u$ to the condition that $1-\frac{4}{st}$ is a rational square. Any oversight here would allow asymmetric solutions.

The decisive restriction $st=4$ used the identity

$b^2(k-4)=a^2k.$

If $k>4$, then both sides force incompatible divisibility: primes dividing $k$ must also divide $k-4$, which is impossible unless $k$ is extremely small. Explicit substitution shows that $k=5,6,7,8$ fail because $(k-4)/k$ equals $1/5,2/6,3/7,4/8$, none of which are squares in $\mathbb{Q}$ since their reduced forms have non-square numerator and denominator.

Finally, equality in $(x+y)(1/x+1/y)\ge 4$ was used; substitution confirms it holds with equality only for $x=y$, so no asymmetric pair survives once $st=4$ is enforced.

Alternative Approaches

A direct quadratic approach starts from

$x,y=\frac{s\pm\sqrt{s^2-\frac{4s}{t}}}{2}$

and studies when the discriminant is a rational square. One can instead homogenize by setting $x=\frac{p}{q}$, $y=\frac{r}{q}$ and converting both conditions into Diophantine equations in integers $p,r,q$. This leads to the same constraint $st=4$ after eliminating variables, but the intermediate algebra is longer and less transparent than the symmetric substitution using $u=\frac{x}{x+y}$.