# Possible values for infinum of polynomial with integer coefficients

I'am trying to find all natural numbers $$a\in\mathbb{N}$$ such that there is polynomial with integer coefficients $$P\in \mathbb{Z}[x]$$ with $$\inf_{x\in\mathbb{R}}\, P(x) = \sqrt{a}.$$

If $$a=b^2$$ for $$b\in\mathbb{N}$$ we obviously can take $$P(x)=x^2 + b$$, but what about $$a$$ that are not perfect squares? I have tried to find any but not succeed.

• Note that if such a polynomial exists, then the infimum is in fact a minimum, and the value of $x\in\Bbb{R}$ for which $P(x)=\sqrt{a}$ is an algebraic number. Commented Nov 16, 2022 at 3:23
• On a tangent, perhaps it is helpful to know that, for example $$\min_{x\in\Bbb{R}}\big(27x^4+a\cdot4x\big)=-a\sqrt[3]{a}.$$ Similar identities exists for higher $n$-th roots, for all odd $n$. Commented Nov 16, 2022 at 3:37
• A near miss: $$\max_{x\in\mathbb{R}}{(-32a^5x^6+6a^2x^2)} = \sqrt{a}.$$ Commented Nov 16, 2022 at 13:49

The polynomial $$P(x) = (2a)^8x^8-(2a)^5x^6-(2a)^5x^4+6a^2x^2+a^2$$ satisfies $$\min_{x\in\mathbb{R}} P(x) = \sqrt{a}.$$

If there is a polynomial of lower degree with the same property, then it must have degree $$6$$ (see below), assuming $$a$$ is not a perfect square.

There is however the degree-$$4$$ polynomial $$R(x) = -(2a)^4x^4+(2a)^2x^3+(2a)^3x^2-3ax-a^2$$ that satisfies $$\max_{x\in\mathbb{R}} R(x) = \sqrt{a},$$ leading to the surprising conclusion that achieving a minimum of $$\sqrt{a}$$ requires a strictly higher degree than achieving a maximum of $$\sqrt{a}$$.

Note that any $$P(x)$$ satisfying the original requirements necessarily has even degree. To see why $$P(x)$$ cannot have degree $$4$$, let $$\beta$$ be one of the real algebraic numbers where $$P(x)$$ achieves its minimum of $$\sqrt{a}$$. Then $$\sqrt{a} = P(\beta)$$ is an element of the field $$\mathbb{Q}(\beta)$$. Moreover, if there is an automorphism $$\sigma$$ of $$\mathbb{Q}(\beta)$$ satisfying $$\sigma(\sqrt{a}) = -\sqrt{a} \qquad \text{and} \qquad \sigma(\beta) \in \mathbb{R},$$ then $$P(\sigma(\beta)) = \sigma(P(\beta)) = \sigma(\sqrt{a}) = -\sqrt{a},$$ meaning $$\sqrt{a}$$ is not actually the minimum of $$P(x)$$. (This is the step where the asymmetry between minimum and maximum becomes apparent.)

In summary, any automorphisms of $$\mathbb{Q}(\beta)$$ that send $$\sqrt{a}$$ to $$-\sqrt{a}$$ must send $$\beta$$ to a non-real number. In particular, $$\beta \not \in \mathbb{Q}(\sqrt{a})$$, so $$\mathbb{Q}(\sqrt{a})$$ is a proper subfield of $$\mathbb{Q}(\beta)$$. Since degrees of field extensions are multiplicative, this means that the degree of $$\mathbb{Q}(\beta)/\mathbb{Q}$$ is even and strictly greater than $$2$$; so $$\beta$$ has degree at least $$4$$. Finally, since $$P'(\beta) = 0$$, $$P(x)$$ must have degree greater than $$4$$.

Here's how I found $$P(x)$$:

The simplest $$\beta$$ that satisfies the above restrictions on $$\mathbb{Q}(\beta)$$ is $$\beta = \sqrt[4]{a}$$. I first constructed a polynomial $$Q(x)$$ with rational coefficients to have minimum $$\sqrt{a}$$ at $$x=\pm\sqrt[4]{a}$$. Here $$Q(x)$$ necessarily has the form $$Q(x) = (x^4-a)q(x)+x^2$$ such that $$q(x)$$ has positive leading coefficient and $$Q'(x)$$ is divisible by $$x^4-a$$. This quickly leads to $$Q(x) = (x^4-a)^2 -\frac{1}{2a}x^2(x^4-a) + x^2.$$ Finally, to get integer coefficients without affecting the minimum, we scale $$Q(x)$$ horizontally; $$P(x) = Q(2ax)$$ does the job.

• Very useful! Can you please explain why $\beta \not \in \mathbb{Q}(\sqrt{a})$ implies that $\beta$ has degree at least $4$? Commented Nov 20, 2022 at 11:24
• Edited to expand that part a bit more. Commented Nov 20, 2022 at 17:52