# a recursion in roots of the polynomial

let $$P(x)$$ is a polynomial which satisfies property $$\psi$$ where property $$\psi$$ is given by

whenever r is a root of $$P(x) = 0$$ then $$r^2 - 4$$ is also a root of the given equation.

i) if $$P(x)$$ is a quadratic polynomial of the form $$x^2 + ax + b$$ then find all the possible equations which satisfy $$\psi$$ and has distinct real roots.

ii) if $$P(x)$$ is a cubic polynomial of the form $$x^3 + ax^2+bx+c$$ then find all the cubic equations which do not share any root with the equations calculated in part (i)

My attempt for part (i)

let $$r$$ is a root of $$P(x)$$ then $$r^2 - 4$$ should also be a root of the equation

and as $$r^2 -4$$ is a root then is $$(r^2-4)^2-4$$ so atleast two of them should be equal otherwise the $$P(x)$$ would be the zero polynomial which is a contradiction so

either $$r = r^2 - 4 \ or \ r = r = (r^2-4)^2-4$$ as the third equation leads to complex roots and again the first is false as the roots should be distinct so $$r$$ satisfies $$(r^2-4)^2-4 = r$$ but I am not able to solve further and also not sure that what i have done is correct I just need the idea to solve the cubic part

by calculator i found four values of r for first part but in the question it was mentioned that there are only 2 polynomials

• CMI-2020 held on 4/10/2020 IST – Anindya Prithvi Oct 4 '20 at 12:51
• yes i also gave it – Aditya Dwivedi Oct 4 '20 at 12:51
• Ohh, I see...I forgot about recursion for the roots. (use sum of roots=-b and product as c) then $D\neq 0$. Use the recursion to prove there exist only two values of r (for part 1) – Anindya Prithvi Oct 4 '20 at 13:01
• @AnindyaPrithvi My main problem is still joined with the text of the question. (Not with 12 marks, whatever they quivalate in dollars or zen.) Which is exactly the framework of the question. Is it so complicated to write $a,b,c\in \Bbb R$ somewhere in the text of the given stuff? Is it so complicated to state immediately after inventing $P$ that $P$ has different real roots? (If this is the case for $P$ for both (i) and (ii).) Just as a further comment on the style of the question. There is no "recursion" (in roots) that can be extracted from the problem. – dan_fulea Oct 4 '20 at 16:58
• @AdityaDwivedi And the text in the OP would / should be improved by just adding the symbols point and comma at the many places where they are missing. And please make clear which is the question. An answer should solve (i) and (ii)? Or just give the roots of $$r=(r^4-4)^2-4\ ?$$ – dan_fulea Oct 4 '20 at 17:00

(i) Let $$a,b$$ be such that $$P(x)=x^2+ax+b$$ has the given property. Let $$x_1,x_2$$ be the roots of $$P(x)$$.

• If $$x_1^2-4=x_2^2-4$$, then we have successively $$x_1^2=x_2^2$$, $$ax_1+b=ax_2+b$$, $$a(x_1-x_2)=0$$, $$a=0$$. So $$x_1,x_2$$ are $$\pm\sqrt {-b}$$, and we need $$-b-4=x_1^2-4=\pm\sqrt b$$. So $$\sqrt b$$ is a root of the equation $$-T^2-4=\pm T$$ in $$T$$. There are four solutions in complex numbers, none of them is real. But i suppose we are working over $$\Bbb R$$. So this case leads to no solution.

• If $$x_1^2-4\ne x_2^2-4$$, the remained case, then these two values are $$x_1,x_2$$ in this order or in the other one. So we have one of the cases: \begin{aligned} x_1 &= x_1^2-4\ ,\\ x_2 &= x_2^2-4\ , \end{aligned} \qquad\text{ or } \begin{aligned} x_1 &= x_2^2-4\ ,\\ x_2 &= x_1^2-4\ . \end{aligned} In the first case we obtain real distinct roots, they are the roots of the equation $$X^2-X-4=0\ ,$$ and we compute them immediately, $$\frac 12(1\pm\sqrt{17})$$. In the second case, we plug in the second equation in the first equation, so $$x_1$$ (as also $$x_2$$ by symmetry) is a solution of $$X=(X^2-4)^2-4\ .$$ (This equation also appears in the OP...) But we have the supplementary information, that also the roots of the first system (obviously) satisfy this equation. So we expect that the polynomial $$(X^2-X-4)$$ divides $$(X^2-4)^2-X-4=X^2-8X^2-X+12$$. This is the case, and the quotient is $$(X^2+X-3)$$. We have thus the two other values, $$\frac 12(-1\pm\sqrt{13})$$.

Well, we do not need the roots (since both sufficient/necessary phases are clear), the two polynomials needed are $$(X^2-X-4)$$ and $$(X^2+X-3)$$.

(ii) is completely unclear, may be complex, may (all three) coincide, etc.

I will solve the explicit case when the roots $$x_1,x_2,x_3$$ are real and distinct. Let $$g(x)$$ be $$g(x)=x^2-4$$. What is $$g(x_1)$$. It cannot be $$x_1$$, else we have a root from (i). Similarly for the other two. So $$g$$ permutes the roots without fixed point. There are only two such permutations. We possibly exchange $$x_2,x_3$$, so that in this case (ii), we have to solve: \tag{*} \left\{ \begin{aligned} x_1 &= g(x_2) = x_2^2-4\ ,\\ x_2 &= g(x_3) = x_3^2-4\ ,\\ x_3 &= g(x_1) = x_1^2-4\ . \end{aligned} \right. We plug in the second, then the third relations in the first one, eliminate $$x_2,x_3$$ and obtain an equation in $$x_1$$. So $$x_1,x_2,x_3$$ are the roots of $$X=g(g(g(X)))$$. Expanding, we get the equation: $$x^8 - 16x^6 + 88x^4 - 192x^2 - x + 140 =0\ .$$ With the same argument from (i), we expect the factor $$(x^2-x-4)$$, but the remained polynomial of degree $$6$$ has not a quick human factorization. Well, using a CAS (computer algebra system) like sage, this is an easy task:

sage: var('x');
sage: g(x) = x^2 - 4
sage: expand( g(g(g(x))) - x )
x^8 - 16*x^6 + 88*x^4 - 192*x^2 - x + 140
sage: factor( g(g(g(x))) - x )
(x^3 + 2*x^2 - 3*x - 5)*(x^3 - x^2 - 6*x + 7)*(x^2 - x - 4)


Here we still have to work a little, since if $$x_1$$ is for instance a root of the first factor $$(x^3 + 2x^2 - 3x - 5)$$, there is a question about $$x_2=g(x_1)=x_1^2-4$$, is it a root of the same factor? The answer is yes, since also using the CAS (sage here):

sage: factor( g(x)^3 + 2*g(x)^2 - 3*g(x) - 5 )
(x^3 + 2*x^2 - 3*x - 5)*(x^3 - 2*x^2 - 3*x + 5)


so $$g(x_1)$$ is a root of the one or the other factor $$(x^3 \pm 2x^2 - 3x \mp 5)$$, but the "bad" factor is excluded, because $$x_1$$ is also a root of the polynomial $$g(g(g(x)))-x$$ above.

But as a human, there is still a good way to proceed, if this happens in an exam with all computer aid excluded. (This exclusion still happens in this century, it's for the protection and the benefit of the students, that else would start playing chess and go during the boring exam on their powerful machines, while also enjoying the hi res albums of past holidays with the family. Sorry...)

To see how, let us start with the system $$(*)$$ above. Let $$s,q,p$$ be the three elementary symmetric expressions of degrees $$1,2,3$$ (Vieta) for $$x_1,x_2,x_3$$, so $$s$$ is their sum, $$p$$ their product, and $$q=x_1x_2+x_2x_3+x_3x_1$$. Then we get relatively quickly some algebraic equations in $$s,q,p$$: \begin{aligned} s^2 &= x_1^2+x_2^2+x_3^2+2q=(x_1+4)+(x_2+4)+(x_3+4)+2q \\ &=s+2q+12\ , \\[2mm] q^2 &= x_1^2x_2^2 + x_1^2x_2^2 + x_1^2x_2^2 + 2ps\\ &=(x_1+4)(x_2+4) + (x_2+4)(x_3+4) + (x_3+4)(x_1+4) + 2ps\\ &=q+8s+48+2ps\ ,\\[2mm] p^2 &=x_1^2x_2^2x_3^2=(x_1+4)(x_2+4)(x_3+4)\\ &=p+4q+16s+64\ ,\\[2mm] 1 &= (x_1+x_2)(x_2+x_3)(x_3+x_1) \\ &=x_1^2(x_2 +x_3) + x_2^2(x_3 +x_1) + x_3^2(x_1 +x_2) + 2x_1x_2x_3\\ &=\underbrace{x_1^2(x_2 +x_3) + x_2^2(x_3 +x_1) + x_3^2(x_1 +x_2) + 3x_1x_2x_3} _{(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)} - x_1x_2x_3\\ &=sq-p\ . \end{aligned} The last equation was obtained starting from $$(x_1-x_2)=(x_2^2-x_3^2)=(x_2-x_3)(x_2+x_3)$$, building the two other cousins obtained cycling the indices, taking the product, and simplifing, since the roots are different.

The last equation, $$p = sq-1$$, can be used to eliminate $$p$$ in the other equations. Then the first two equations become $$q = \frac 12(s^2-s-12)\ ,\ 0=25q+6s+48+2qs+3q^2\ .$$ We substitute $$q$$ from the first equation in the second equation. The equation in $$s$$ is $$3s^4-2s^3-23s^2-2s+24=0\ .$$ The solutions are $$1,3,-2,-4/3$$.

The corresponding $$q$$ values are respectively $$-6,-3,-3,-40/9$$.

The corresponding $$p=sq-1$$ values are respectively $$-7,-10,5,133/27$$.

But only two of them also satisfy the equation $$p^2=p+4q+16s+64$$. The two solutions for $$(s,q,p)$$ are thus:

$$(1,-6,-7)\text{ and }(-2,-3,5)\ .$$ The corresponding polynomials are: $$x^3-x^2-6x+7\text{ and }x^3+2x^2-3x-5\ .$$ $$\square$$

Part i):

Let $$r, s$$ be the two roots of $$P$$.

By assumption, $$r^2 - 4$$ and $$s^2 - 4$$ all belong to the set $$\{r, s\}$$. Thus there are several possibilities:

1. $$r^2 - 4 = r$$, $$s^2 - 4 = s$$.

In this case, $$r$$ and $$s$$ are the two roots of $$x^2 - x - 4$$, and we have $$P(x) = x^2 - x - 4$$.

2. $$r^2 - 4 = r$$, $$s^2 - 4 = r$$.

In this case, $$r$$ is one of the two roots of $$x^2 - x - 4$$, which are $$\frac {1 \pm \sqrt{17}}2$$, and since $$s^2 = 4 + r = r^2$$, we have $$s = - r$$.

Therefore $$P(x) = (x - r)(x + r) = x^2 - r^2 = x^2 - \frac{9 \pm \sqrt{17}}2$$.

3. $$r^2 - 4 = s$$, $$s^2 - 4 = r$$.

In this case, we have $$(r^2 - 4)^2 - 4 = r$$, which gives $$(r^2 - r - 4)(r^2 + r - 3) = 0$$.

Since $$r\neq s$$, we have $$r^2 - r - 4 \neq 0$$, so $$r$$ is a root of the polynomial $$x^2 + x - 3$$.

The same argument shows that $$s$$ is also a root of $$x^2 + x - 3$$. Therefore $$P(x) = x^2 + x - 3$$.

Part ii):

We now have three roots $$r, s, t$$ (which are a priori not necessarily real numbers).

The number $$r^2 - 4$$ cannot be equal to $$r$$, so it is one of $$s$$ and $$t$$, let's say $$s$$.

And $$s^2 - 4$$ cannot be equal to neither $$s$$ nor $$r$$ (otherwise $$s$$ satisfies $$(s^2 - 4)^2 - 4 = s$$ and coincides with one of the previous roots), so it must be $$t$$.

Finally, $$t^2 - 4$$ cannot be equal to neither $$t$$ nor $$s$$, for the same reason as above, so it must be $$r$$.

Therefore we have $$((r^2 - 4)^2 - 4)^2 - 4 = r$$. After factorization, we get: $$(r^2 - r - 4) (r^3 - r^2 - 6r + 7) (r^3 + 2r^2 - 3r - 5) = 0.$$ The first factor contains only previous roots, so we see that $$r^3 - r^2 - 6r + 7 = 0$$ or $$r^3 + 2r^2 - 3r - 5 = 0$$.

If $$r^3 - r^2 - 6r + 7 = 0$$, then $$s = r^2 - 4$$ also satisfies $$s^3 - s^2 - 6s + 7 = 0$$, as can be verifies as follows: $$(r^2 - 4)^3 - (r^2 - 4)^2 - 6(r^2 - 4) + 7 = (r^3 - r^2 - 6r + 7) (r^3 + r^2 - 6r - 7) = 0.$$

For the same reason, $$t$$ also satisfies $$t^3 - t^2 - 6t + 7 = 0$$. Therefore $$r, s, t$$ are exactly the three different roots of the polynomial $$x^3 - x^2 - 6x + 7$$, so that $$P(x) = x^3 - x^2 - 6x + 7$$.

Similarly, if $$r^3 + 2r^2 - 3r - 5 = 0$$, then we conclude that $$P(x) = x^3 + 2x^2 - 3x - 5$$.