Does there exist a polynomial over integers possesing certain property?

Question

Does there exist a polynomial with integer coefficients which posseses the local minimal value $\sqrt{2}$ (not a local minimum at $\sqrt{2}$)?

Answer 1

$$4x^4-2x^3-4x^2+3x+1$$ has a local minimum at $x=\frac{\sqrt2}2$ with value $\sqrt2$.

So agrees Wolfram.

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How did I find this? I first chose what the $x$-value would be. It turned out that $x_0=\sqrt2$ would lead to non-integer coefficients, so I changed it to $\sqrt2/2$. Then I looked for a low-degree polynomial which would have a stationary point with $f(\sqrt2/2)=\sqrt2$.

With integer coefficients $f(x_0)$ and $f'(x_0)$ would always be of the form $P+Q\sqrt2$. The even coefficients influence $P$ of $f(x_0)$ and $Q$ of $f'(x_0)$ which must both vanish, so lets just look at the odd coefficients, which influence $Q$ of $f(x_0)$ and $P$ of $f'(x_0)$, which must be 1 and 0, respectively. The simplest possibility would be that there are first and third degree coefficients, so let's solve $$\begin{array}{rll} a_3x_0^3 + a_1x_0 &= \tfrac{\sqrt2}4a_3 + \tfrac{\sqrt2}2a_1 &= \sqrt2 \\ 3a_3x_0^2 + a_1 &= \tfrac32a_3 + a_1 &= 0 \end{array}$$ which gives $a_3=-2$ and $a_1=3$ -- fortunately both integers. So

$$ f_0(x) = -2x^3+3x $$ has $f_0(x_0)=\sqrt2$ and $f_0'(x_0)=0$. Unfortunately it has $\sqrt2$ as a local maximum.

In order to correct for that I look for a polynomial with a double root at $x_0$ which I can add a suitable multiple of to make the local maximum into a local minimum. A good candidate for this would be $$ \left(x-\frac{\sqrt2}2\right)^2\left(x+\frac{\sqrt2}2\right)^2 = x^4-x^2+1/4 $$ except I need to multiply it by $4$ to get integer coefficients. And it turns out that adding $4x^4-4x^2+1$ is enough to swap the local maximum to a minimum.