AoPS - Problem

Source: China TST 2002 Quiz

Tags: number theory unsolved, number theory

shobber

28.01.2009 19:07

For positive integers $a,b,c$ let $ \alpha, \beta, \gamma$ be pairwise distinct positive integers such that \[ \begin{cases}{c} \displaystyle a &= \alpha + \beta + \gamma, \\ b &= \alpha \cdot \beta + \beta \cdot \gamma + \gamma \cdot \alpha, \\ c^2 &= \alpha\beta\gamma. \end{cases} \]Also, let $ \lambda$ be a real number that satisfies the condition \[\lambda^4 -2a\lambda^2 + 8c\lambda + a^2 - 4b = 0.\]Prove that $\lambda$ is an integer if and only if $\alpha, \beta, \gamma$ are all perfect squares.

It is easy to check the equation has four roots as follows: $ \sqrt{\alpha}+\sqrt{\beta}-\sqrt{\gamma},\sqrt{\alpha}-\sqrt{\beta}+\sqrt{\gamma},-\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma},-\sqrt{\alpha}-\sqrt{\beta}-\sqrt{\gamma}$. Now sufficiency is trivial. To prove necessity, we assume, WLOG, $ \sqrt{\alpha}+\sqrt{\beta}-\sqrt{\gamma}=\lambda$. We can square $ \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\gamma}+\lambda$ and see $ \sqrt{\alpha\beta}-\lambda\sqrt{\gamma}$ is rational. Similarly, we see $ \sqrt{\beta\gamma}-\lambda\sqrt{\alpha},\sqrt{\alpha\gamma}-\lambda\sqrt{\beta},\sqrt{\alpha\beta}-\lambda\sqrt{\gamma}$ are rationals. Since $ c=\sqrt{\alpha\beta\gamma}$ is integer, we see $ \sqrt{\alpha\beta\gamma}-\lambda\alpha,\sqrt{\alpha\beta\gamma}-\lambda\beta,\sqrt{\alpha\beta\gamma}-\lambda\gamma$ are integers. Now it is trivial to prove $ \sqrt{\alpha},\sqrt{\beta},\sqrt{\gamma}$ are all rationals. Since $ \alpha,\beta,\gamma$ are integers, we know they are also perfect squares. Q.E.D.