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Please help to solve this question:

polynomial with rational coefficients have $-\sqrt{2}$ as it minimal value ?

Thanks allot!

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It all depends: if you allow only rational values of $\,x\,$ then the answer is clearly no, but it you allow all realsm or even all complex, values for $\,x\,$ then perhaps things can change... – DonAntonio Jan 12 '13 at 16:54
@DonAntonio If you allow for complex values of $x$, there is no minimum (except for constant functions) – Calvin Lin Jan 12 '13 at 16:58
I would consider it 'cheating', but if you are allowed to restrict the domain, then the answer is very easily yes. Presumably you want a polynomial defined over $\Bbb R$ though. – Clayton Jan 12 '13 at 17:00
Yes, @CalvinLin, that's part of the "perhaps things can change" phrase above. This follows at once from Liouville's Theorem. – DonAntonio Jan 12 '13 at 17:01

The polynomial $\frac14(\frac14 x^4-x^3-x^2+6x+1)$ has $-\sqrt2$ as a minimum at $x=-\sqrt2$.

This is I found it: The minimum needs to be attained at an irrational value, so I set out to find a polynomial where it is attained at $\pm\sqrt2$. Thus I wanted $x^2-2$ to be a factor of $f'$, but I also want $f'$ to have an odd degree so that $f$ has a minimum. Thus I tried $f'(x)=\alpha(x^2-2)(x-q)$ for some rational $\alpha,q$, integrated and added an arbitrary rational constant $r$. Choosing appropriate constants yields the desired minimum at $-\sqrt2$.

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What was the motivation behind the construction? Clearly we need at least 4th power. – Calvin Lin Jan 12 '13 at 17:18
Integrate $\alpha(x^2-2)(x-q)$ and add $r$. Then insert $x=-\sqrt2$ into that function. You get $-4/3 \sqrt2 q \alpha+r-\alpha$. This will equal $-\sqrt2$ if you choose any values of $q,r,\alpha$ such that $r=\alpha$ and $q\alpha=3/4$, for example $r=\alpha=1/4, q=3$. – Samuel Jan 15 '13 at 7:27

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