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Consider polynomials $p(x)$ with the following property, called $(†)$. $(†)$: If r is a root of $p(x)$, then $r^2 − 4$ is also a root of $p(x)$. We want to find every quadratic polynomial of the form $p(x) = x^2 +bx+c$ such that $p(x)$ has two distinct roots, has integer coefficients, and has property $(†)$. Prove that there are exactly two such polynomials and list them.

If $r^2-4$ is a root then it would imply $(r^2-4)^2-4$ is a root which would again imply $((r^2-4)^2-4)^2-4$ is a root and so on. But since we can have only two roots, we must have all the roots except $r$ and $r^2-4$ to be equal to either of them. Is my reasoning correct? How to explicitly find the equations?

Tapi
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    Note that the polynomial is of degree 2, so it has at most two roots. So $(r^2-4)^2-4$ is either equal to $r$ or to $r^2-4$. – acupoftea Nov 29 '20 at 19:55
  • Once you have the desired value of $r$ per @acupoftea, it is easy to construct the quadratic polynomial with those "two" roots. Another possibility is that $r^2 - 4 = r$. – hardmath Nov 29 '20 at 19:58
  • @acupoftea do I not need to account for the rest like $((r^2-4)^2-4)^2-4$? – Tapi Nov 29 '20 at 20:27
  • @hardmath It's given that the roots are distinct, so will that work? – Tapi Nov 29 '20 at 20:28
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    @Tapi The polynomial either has one root or two, if it's only one, call it $p$, we know that $p^2-4=p$, because $p^2-4$ has to be a root, but there's only one. So you can write and solve that equation. If there are two roots, then by applying the function $f(x)=x^2-4$ to either of them you must get either the same one that you started with or the other one by the same reasoning. – acupoftea Nov 29 '20 at 20:32
  • Vieta's formulas highlight the links between the roots and coefficients of a polynomial. – rtybase Nov 29 '20 at 20:51
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    @Tapi: Yes, since $r^2 - r - 4 = 0$ has two distinct real roots (positive discriminant). The coefficients have to be integers, but the roots are not necessarily integers. – hardmath Nov 29 '20 at 21:06
  • See a recursion in roots of the polynomial – mathlove Nov 30 '20 at 06:20

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$$p(x)=x^2- x-4$$ indeed solving the equation we get $$x_1=\frac{1}{2} \left(1-\sqrt{17}\right);\;x_2=\frac{1}{2} \left(\sqrt{17}+1\right)$$ $$\left(\frac{1}{2} \left(1-\sqrt{17}\right)\right)^2-4=\frac{1}{2} \left(1-\sqrt{17}\right)$$ $$\left(\frac{1}{2} \left(1+\sqrt{17}\right)\right)^2-4=\frac{1}{2} \left(\sqrt{17}+1\right)$$

Raffaele
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