<s>[INST] Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$ Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$.
It it a high-school math contest problem. The answer is given:
$$(a) ~~~~~~~P(x)=N(2x^2-1)^2-2x^3+3x,~~~~N>1$$
$$(b) ~~~~~~~Q(x)=P(x^2)=N(2x^4-1)^2-2x^6+3x^2$$
No solution is given.
My initial approach was (before I saw the answer) to try finding the minimun of a general polynomial starting with degree $3$ and then try to match the coefficients to a given value of the minimum. However, the answer clearly shows that we require degrees $4$ and $8$ respectively, which makes my method of solution practically impossilbe.

It's easy enough to show that the answers are true, for example the case (a):
$$P'(x)=8N(2x^2-1)x-6x^2+3=(2x^2-1)(8Nx-2)=0$$
$$x_1=\frac{1}{\sqrt{2}},~~~~~~~x_2=-\frac{1}{\sqrt{2}},~~~~~~~x_3=\frac{1}{4N}$$
$$P''(x)=8N(2x^2-1)+4x(8Nx-2)$$
$$P''(x_1)=2\sqrt{2}(4N\sqrt{2}-2)>0$$
$$P''(x_2)=-2\sqrt{2}(-4N\sqrt{2}-2)>0$$
$$P''(x_3)=\frac{1}{N}-8N<0$$
So, $x_1$ and $x_2$ are minimum points, $x_3$ is a maximum point.
$$P(x_1)=\frac{3\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=\sqrt{2}$$
$$P(x_2)=-\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=-\sqrt{2}$$

The case (b) follows trivially, if we replace $x \to x^2$.


How is this problem supposed to be solved? How are we supposed to find the degree of $P(x)$ and match the coefficients? Is there some theorem about a minimal value of a polynomial which might help?

 [/INST] For part (a), it can be easily deduced that $f$ does not have degree 2. Further, $f$ cannot have degree 3, as an odd degree polynomial has no global minimum. So assume 
$$f(x) = sx^4 + bx^3 + cx^2 + dx + e$$
with $f(a) = -\sqrt{2}, f'(a) = 0$. As one last piece of guesswork, let $a = k\sqrt{2}$.  Then we have
$$4sk^4 + 2bk^3\sqrt{2}+2ck^2+dk\sqrt{2}+e = -\sqrt{2}$$
$$8sk^3\sqrt{2}+6bk^2+2ck\sqrt{2}+d=0$$
from which we derive
$$4sk^4 + 2ck^2 + e = 0$$ 
$$2bk^3 + dk = -1$$
$$8sk^3 + 2ck = 0$$
$$6bk^2 + d = 0$$
Then $c = -4sk^2, e = 4sk^4, d = -1.5/k, b = 0.25/k^3$. Now we just need to find $k, s$ that makes all of these integers. $k = 1/2, s = 4$ does the trick. Then we have 
$$f(x) = 4x^4 + 2x^3 - 4x^2 - 3x + 1$$
Which has all of our desired properties. 
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