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Consider the polynomials $x^3+ax^2+c$ and $x^3+cx^2+a$ where $a,c\in\mathbb Z$. Is it possible that both of the polynomials have three distinct integer roots? If yes, find such $a$ and $c$ such that $\lvert a\rvert+\lvert c\rvert$ is minimum.

I'm intuitively thinking that such $a$ and $c$ exists. But I don't have a good reason to convince this to someone. I've done the following:

Let the roots of the first polynomial $x^3+ax^2+c$ be $R_1,R_2,R_3$. By Vieta's formula, we have $$\begin{align} & R_1+R_2+R_3=-a\\ & R_1R_2+R_2R_3+R_1R_3=0\\ & R_1R_2R_3=-c\end{align}$$ Let the roots of the second polynomial $x^3+cx^2+a$ be $r_1,r_2,r_3$. By Vieta's formula, we have $$\begin{align} & r_1+r_2+r_3=-c\\ & r_1r_2+r_2r_3+r_1r_3=0\\ & r_1r_2r_3=-a\end{align}$$ So we have $$\begin{align} & R_1+R_2+R_3=r_1r_2r_3\\ & R_1R_2+R_2R_3+R_1R_3=r_1r_2+r_2r_3+r_1r_3\\ & R_1R_2R_3=r_1+r_2+r_3\end{align}$$ I'm unable to proceed from here.

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    $\begingroup$ @user170231 Doesn't the RRT allow for the roots to be factors of $c$ or their additive inverses as well? $c$ is not indicated as prime in the OP. $\endgroup$ Oct 15, 2021 at 1:08
  • $\begingroup$ The sum of three numbers equaling their product might ring a bell. There are not many solutions because if two of them are large the product is larger yet. $\endgroup$ Oct 15, 2021 at 4:29

1 Answer 1

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For faster typing, I'm going to label the roots of the first equation $\{m, n, p\}$, and of the second equation $\{t,u,v\}$. So we have:

$$ \begin{gather} m+n+p = tuv = -a \\ mnp = t+u+v = -c \\ mn+mp+np=tu+tv+uv = 0 \end{gather} $$

A lot of the following requires assuming WLOG, as $\{m, n, p\}$ and $\{t, u, v\}$ are all interchangeable within their own sets, and the sets themselves are swappable.

We can see that $a, c \neq 0$. If $a=0$, then at least one of $t,u,v = 0$. Choose $t=0$. Then

$$tu+tv+uv = 0u+0v+uv = 0 \implies uv=0$$

That means $t = u = 0$, which doesn't give us three distinct roots. The same applies to $c$. Therefore, each of $a,c$ must have three different (not necessarily prime) factors.

To have a solution, we need two distinct sets of three distinct integers. Each of them must have a sum equal to the other's product, and the sum of the pairwise products must be equal to zero. That last bit means each set must have at least one negative and at least one positive integer. For the sake of simplicity, we will choose $m>0, p<0$ and $t>0, v < 0$.

Therefore, we need two sets such that $mnp = t+u+v$ and $tuv = m+n+p$. To have that, we require:

$$|mnp| > |m+n+p| \iff |tuv| < |t+u+v|$$

(The values cannot be equal, or our two sets will be the same.) And here's our wall: there are no three distinct integers $t, u, v$ such that $|t+u+v| > |tuv|$. There are cases where they are equal, e.g., $1+2+3 = 1 \cdot 2 \cdot 3$ and $|(-1)(1)(k)| = |k+1-1|$, but none where the sum is greater.

So, alas, your intuition was incorrect. As was mine, as I thought I'd find at least one solution. Hope this helps!

(Brief side note: if we don't require distinct integers, $\{1, 1, k\}$ and $\{1,2,2\}$ have sums less than their product. I believe those are the only sets that work, though.)

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  • $\begingroup$ Great answer! Thanks. $\endgroup$
    – Oshawott
    Oct 15, 2021 at 16:41

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