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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?

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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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  • $\begingroup$ Thanks. I didn't think why degree 3 doesn't work. This seems like the best way, since guesswork is widely used in math contests $\endgroup$ – Yuriy S Jul 4 '16 at 13:09
  • $\begingroup$ @YuriyS IMO this is one of the problems that you'd want to skip. $\endgroup$ – florence Jul 4 '16 at 13:10
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This like a "find the minimum of a polynomial" problem you've seen countless times, but you're being asked to work backwards from the answer to the question!

The question tells you that the minimums of $f$ satisfy $(f-\sqrt{2})(f+\sqrt{2})=0$

You know that at the minimum, the gradient is zero so you might try $f'=(f-\sqrt{2})(f+\sqrt{2})$.

Then be mindful that you may in due course have to choose some arbitrary parameter(s) to ensure that the points are minimums rather than maximums and there is no other minimum.

Finally you may need to apply some algebra to the coefficients ensure they are all integers.

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  • $\begingroup$ Just a suggestion but ...let $f=y$ for now. Then $f'$ means $\frac{dy}{dx}$. $\endgroup$ – user334732 Jul 4 '16 at 11:13
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    $\begingroup$ Ah, I see. It's not an ODE because we only consider $f(x_0)$ and $f'(x_0)$ for $x_0$ a minimum point. Still, this is a multi-variable equation. And how are we supposed to guess the degree of $f$? $\endgroup$ – Yuriy S Jul 4 '16 at 11:22
  • $\begingroup$ I may have misunderstood because I thought this was the entire question: Find a polynomial with integer coefficients which has a global minimum equal to (a)−2‾√ 2 , (b)2‾√ 2 . $\endgroup$ – user334732 Jul 4 '16 at 11:34
  • $\begingroup$ Ok I don't understand why it says "The answer is given:" and it also says "No solution is given". This seems a contradiction to me. $\endgroup$ – user334732 Jul 4 '16 at 11:51
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    $\begingroup$ It means that there is no information on how this answer was obtained $\endgroup$ – Yuriy S Jul 4 '16 at 11:54

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