Competition problem — Number theory

Library / /327 of 520

Number theory Difficulty 5.8 AIME, harder Prove it

$\underline{\text { Agakhanov N.K. }}$

Real numbers $x$ and $y$ are such that for any distinct odd primes $p$ and $q$ , the number $x^{p}+y^{q}$ is rational.

Prove that $x$ and $y$ are rational numbers.

Solution

From the rationality of $x^{p}+y^{q}, x^{r}+y^{q}, x^{s}+y^{q}$ follows the rationality of the numbers $x^{r}-x^{p}, x^{s}-x^{r}$ . Let $p=3, r=$ 5, $s=7$ . Then $a=x^{7}-x^{5}$ and $b=x^{5}-x^{3}$ are rational. If $b=0$ , then $x=0$ or $x= \pm 1$ , that is, $x$ is rational. If $b \neq 0$ , then $x^{2}=a / b$ is rational. But then from the equality $b=x^{2}\left(x^{2}-1\right) x$ follows the rationality of $x$ . Similarly, $y$ is a rational number.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.