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Problem. Let $P$ be a polynomial with integer coefficients and consider $m(P) = \min_\limits{x\in\mathbb{R}} P(x)$. The problem is to describe all possible values of $m(P)$.

Necessary condition. If $m(P)$ is attained at the point $x_0$ then $P'(x_0) = 0$ hence $x_0$ is algebraic and so is $m(P) = P(x_0)$. Thus $m(P)$ has to be algebraic, but this is the only restriction I was able to establish.

Sufficient conditions. Some explicit examples can be constructed.

Let $q$ be a positive integer. The minimal value of the polynomial $P_q(x) = x^2q - 2x$ equals $-1/q$ therefore the minimal value of a polynimal of the form $rP_q(x) + p$ for integer $p,r > 0$ can be any rational value with denominator $q$. It follows that $m(P)$ can be any rational number.

For the situation when $m(P)$ is irrational there is a well-known question for $\sqrt{2}$, e.g., see this question. Following the same idea one can show that $m(P_{a, N}) = \sqrt{a}$, for $$ P_{a,N} = N(16a^2x^4 - 8ax^2 + 1) - (4a^2x^3 - 3ax) $$ where $a$ and $N$ are positive integers and $N$ is large enough. This method can be extended to construct examples for $m(P) = -\sqrt{a}$ and (I did not check this part, but it seems provable) to the values of the form $q\sqrt{a}$ with rational $q$.

Thoughts. Denote the set of all possible values of $m$ by $M$.

  • Does $M$ coincide with the set of algebraic numbers?
  • Is $M$ closed under addition?
  • The same question for multiplication and division.
  • $M$ is clearly closed under the multiplication by positive integers. Is it closed also under division by them? Positive answer will imply that to define $M$ one can consider polynomials not only in $\mathbb{Z[x]}$ but in $\mathbb{Q[x]}$.
Pavel Gubkin
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  • Clearly if the degree of $P$ is odd or the leading coefficient is negative, then $P$ has no minimum. How does that fit into this? Are you limiting to polynomials with a minimum, or should $-\infty$ be considered a member of $M$? – Paul Sinclair Dec 14 '22 at 19:12
  • Yes, I think the best way is to restrict the problem to polynomials of even degree with positive leading coefficient. Number $-\infty$ creates unnecessary problems concerning multiplication – Pavel Gubkin Dec 15 '22 at 11:12

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