Let $p\le q$ be roots of the (real) quadratic equation $x^2+ax+b=0$, $|p|+|q|\ne 0.$ Form the new equation $x^2+px+q=0$, find its real roots (if exist), etc. For example, if $a=3, b=2$, then $p=-2,q=-1$, $x^2-2x-1=0$ has two real roots $p_1= 1 - \sqrt{2}$ and $q_1= 1 + \sqrt{2}$ (note that $p_1\le q_1$) but the equation $x^2+p_1x+q_1$ does not have real roots, the process ends.

Question: What is the longest possible sequence of quadratic equations we can get?

  • 1
    $\begingroup$ If it's possible to make something like $p$ and $q$ keep switching back and forth between 2 values, does that count as $2$ or $\infty$? $\endgroup$ Jun 9, 2020 at 19:49
  • $\begingroup$ @user6247850 I think it would be infinity (OP please confirm). Did you find such an example? $\endgroup$
    – paulinho
    Jun 9, 2020 at 19:52
  • $\begingroup$ Infinity is not possible. The process always terminats after finitely many steps. $\endgroup$
    – markvs
    Jun 9, 2020 at 23:15
  • $\begingroup$ @user158834 how do you see it always terminates? $\endgroup$
    – doetoe
    Jun 9, 2020 at 23:50
  • $\begingroup$ @doetoe: It is basically obvious. The key is that $p_i\le q_i$. $\endgroup$
    – markvs
    Jun 10, 2020 at 0:12

2 Answers 2


Let's work backwards: given a final state, how many steps would it take to reach the beginning? Let the final state be represented by $r_1, r_2$ with $r_1 \le r_2$.

We can see that the quadratic before the final one must have been $$(x-r_1)(x-r_2) = x^2+(-r_1-r_2)x + r_1r_2$$

In order to have a sequence longer than $1$, it must be true that $$-r_1-r_2 \le r_1r_2 \tag 1$$

To get that quadratic, the previous quadratic must have been $$(x+r_1+r_2)(x-r_1r_2) = x^2 + (r_1+r_2-r_1r_2)x - r_1r_2(r_1+r_2)$$

In order to get a sequence longer than $2$, it must be true that $$r_1+r_2-r_1r_2 \le - r_1r_2(r_1+r_2) \tag 2$$

Taking it one more step, the following condition must also hold in order to get a sequence longer than $3$: $$(r_1r_2-r_1-r_2+r_1r_2(r_1+r_2)) \le (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)) \tag 3$$

Finally, this last condition must also hold in order to get a sequence longer than $4$: $$-\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)-\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\le\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right) \tag 4$$

In order to satisfy $(2)$, it must be true that $r_1 < 0$ or that $r_2 < 0$. Letting the $x$-axis be $r_1$ and the $y$-axid be $r_2$, this eliminates the first quadrant. In order to satisfy $(3)$, it must be true that $r_1 > 0$ or that $r_2 > 0$, eliminating the third quadrant. However, in order to satisfy both $(1)$ and $(4)$, $(r_1, r_2)$ can only be in the second and fourth quadrants. Therefore, there are no real $r_1, r_2$ that satisfy $(1), (2), (3), (4)$.

This then means that the maximum length of a sequence of the quadratic equations is $4$, obtained by setting $$a = r_1r_2-r_1-r_2+r_1r_2(r_1+r_2), b = (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)$$

for any $r_1, r_2$ that satisfy $(1), (2), (3)$, and $r_1 \le r_2$.

Edit: The conditions $(1), (2), (3), r_1 \le r_2$ can be rewritten as $$-\frac{r_{2}}{r_{2}+1}\le r_{1}\le\frac{-\left(r_{2}^{2}+1-r_{2}\right)+\sqrt{\left(r_{2}^{2}+1-r_{2}\right)^{2}-4r_{2}^{2}}}{2r_{2}}$$ with $r_1 \le 0 \le r_2$

  • $\begingroup$ That is correct. $\endgroup$
    – markvs
    Jun 10, 2020 at 4:57

The equation $y=x^2 + px + q$ can be rewritten as $$y-\left(q-\frac{p^2}{4}\right)=\left(x-\left(\frac{-p}{2}\right)\right)^2$$ And so it is clear that this parabola's vertex is at the point $(\frac{-p}{2},q-\frac{p^2}{4})$. Therefore the process will terminate at step $n$ if $q_n-\frac{{p_n}^2}{4} > 0$.

  • $\begingroup$ This is just a good first step. I suspect there is no limit to the length of a sequence we can create. $\endgroup$
    – K.defaoite
    Jun 9, 2020 at 23:36
  • 2
    $\begingroup$ This is equivalent to the discriminant being non-negative, no? $\endgroup$
    – paulinho
    Jun 10, 2020 at 0:13
  • $\begingroup$ This is not a good first step. See the correct sollution. $\endgroup$
    – markvs
    Jun 10, 2020 at 5:41

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