2

Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.

My Approach:

I could not find any elegant method that is why I tried applying quadratic formula: the roots will be $\frac{-p±\sqrt{p^2-4q}}{2}$ Now we have 2 cases:

  1. $$\left(\frac{-p+\sqrt{p^2-4q}}{2}\right) =\left (\frac{-p-\sqrt{p^2-4q}}{2}\right)^2 -2$$
  2. $$\left(\frac{-p-\sqrt{p^2-4q}}{2}\right)=\left(\frac{-p+\sqrt{p^2-4q}}{2}\right)^2-2$$

It's very tedious to solve these. I searched on Wolfram alpha and got these I II

Is there an elegant way to approach this problem?

user961447
  • 319
  • May be without solving the equation. Notice that : If $\alpha$ and $\beta$ are the roots of $x^2 + p x + q = 0$ then : $$\left{\begin{array}{lcl} \alpha + \beta & = & -p \[3mm] \alpha , \beta & = & q \end{array}\right.$$ – Essaidi Feb 09 '22 at 12:00

2 Answers2

7

Let $\alpha, \beta$ be roots of this polynomial. If $\alpha=\beta$, then $\alpha^2-2=\alpha$, $\alpha=-1, 2$, and then $(p,q)=(2,1)$ or $(-4, 4)$, both choices work.

Suppose $\alpha\ne \beta$.

If $\alpha^2-2=\beta$, $\beta^2-2=\alpha$, then $(\alpha^2-2)^2-2=\alpha$, $\alpha^4-4\alpha^2-\alpha+2=0$, same for $\beta$. The polynomial $x^4-4x^2-x+2$ is $(x - 2)(x + 1)(x^2+x-1)$, $\alpha,\beta\in\{2, -1\}$ do not work. So $\{\alpha,\beta\}=\{(-1+\sqrt{5})/2, (-1-\sqrt{5})/2\}$. It works, $(p,q)=(1,-1)$.

Finally it could be that $\alpha^2-2=\alpha$ which means $\alpha\in\{-1,2\}$, and $\beta^2-2=\alpha$, so either $\alpha=-1$, $\beta^2-2=-1$, $\beta^2=1$, $\beta=1$ since $\beta\ne \alpha$, $(p,q)=(0,-1)$ or $\alpha=2$, $\beta^2-2=2$, $\beta=-2$, $(p,q)=(0,-4)$

Hence the answer: $(p,q)\in\{(-4,4), (2,1), (1,-1), (0,-1), (0,-4)\}$.

markvs
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    Does the question mean that $\alpha$ , $\beta$, $\alpha^2-2$ are the roots of $x^2 +px+q=0$? But since the equation is of degree $2$, it cannot have $3$ roots. Moreover, if we consider $(p,q) = (-418,7960)$ then the roots are $20$ and $398$ which satisfies the condition as $398 = 20^2 -2$. – user961447 Feb 11 '22 at 14:38
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    You should read the question. $398^ 2-2 \ne 20$. – markvs Feb 11 '22 at 15:11
  • @markvs actually given answer is $6$ – mathophile Mar 06 '23 at 18:38
2

For each $\alpha \in \mathbb{R}$ we can define a quadratic with zeros $x_1 = \alpha $ and $x_2 = \alpha^2 - 2$ by using Vieta's Theorem, which states $x^2+px+q$ with $-p = x_1 + x_2$ and $q = x_1 x_2$ has the desired zeros $x_1,x_2$. Hence for a given $\alpha$ the quadratic you are looking for is $x^2 - (\alpha^2 + \alpha - 2)x + (\alpha^3 - 2\alpha)$.

Edit: In order to find solutions s.t. it works indepently from the choice of the root, which we call $\alpha$, the both roots must satisfy $x_1^2 -2 = x_2$ and $x_2^2-2 = x_1$. How to go on from there is already shown in the answer by markvs.

includeCMath
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