
Problem 2. Find all pairs of integers $(a, b)$ so that each of the two cubic polynomials

$$
x^{3}+a x+b \text { and } x^{3}+b x+a
$$

has all the roots to be integers.


Solution. The only such pair is $(0,0)$, which clearly works. To prove this is the only one, let us prove an auxiliary result first.

Lemma If $\alpha, \beta, \gamma$ are reals so that $\alpha+\beta+\gamma=0$ and $|\alpha|,|\beta|,|\gamma| \geq 2$, then

$$
|\alpha \beta+\beta \gamma+\gamma \alpha|0$. Then $\gamma=$ $-(\alpha+\beta)$, so by substituting this,

$$
|\alpha \beta+\beta \gamma+\gamma \alpha|=\left|\alpha^{2}+\beta^{2}+\alpha \beta\right|,|\alpha \beta \gamma|=|\alpha \beta(\alpha+\beta)|
$$

So we simply need to show $|\alpha \beta(\alpha+\beta)|>\left|\alpha^{2}+\beta^{2}+\alpha \beta\right|$. Since $|\alpha| \geq 2$ and $|\beta| \geq 2$, we have

$$
\begin{aligned}
& |\alpha \beta(\alpha+\beta)|=|\alpha||\beta(\alpha+\beta)| \geq 2|\beta(\alpha+\beta)| \\
& |\alpha \beta(\alpha+\beta)|=|\beta||\alpha(\alpha+\beta)| \geq 2|\alpha(\alpha+\beta)| .
\end{aligned}
$$

Adding these and using triangle inequality,

$$
\begin{aligned}
2|\alpha \beta(\alpha+\beta)| & \geq 2|\beta(\alpha+\beta)|+2|\alpha(\alpha+\beta)| \geq 2|\beta(\alpha+\beta)+\alpha(\alpha+\beta)| \\
& \geq 2\left(\alpha^{2}+\beta^{2}+2 \alpha \beta\right)>2\left(\alpha^{2}+\beta^{2}+\alpha \beta\right) \\
& =2\left|\alpha^{2}+\beta^{2}+\alpha \beta\right|
\end{aligned}
$$

Here we have used the fact that $\alpha^{2}+\beta^{2}+2 \alpha \beta=(\alpha+\beta)^{2}$ and $\alpha^{2}+\beta^{2}+\alpha \beta=\left(\alpha+\frac{\beta}{2}\right)^{2}+\frac{3 \beta^{2}}{4}$ are both nonnegative. This proves our claim.

For our main problem, suppose the roots of $x^{3}+a x+b$ are the integers $r_{1}, r_{2}, r_{3}$ and the roots of $x^{3}+b x+a$ are the integers $s_{1}, s_{2}, s_{3}$. By Vieta's relations, we have

$$
\begin{gathered}
r_{1}+r_{2}+r_{3}=0=s_{1}+s_{2}+s_{3} \\
r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}=a=-s_{1} s_{2} s_{3} \\
s_{1} s_{2}+s_{2} s_{3}+s_{3} s_{1}=b=-r_{1} r_{2} r_{3}
\end{gathered}
$$

If all six of these roots had an absolute value of at least 2 , by our lemma, we would have

$$
|b|=\left|s_{1} s_{2}+s_{2} s_{3}+s_{3} s_{1}\right|<\left|s_{1} s_{2} s_{3}\right|=\left|r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}\right|<\left|r_{1} r_{2} r_{3}\right|=|b|
$$

which is absurd. Thus at least one of them is in the set $\{0,1,-1\}$; WLOG, suppose it's $r_{1}$.
