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I am dealing with the test of the OBM (Brasilian Math Olimpyad), University level, 2017, fase 2.

I hope someone can help me discussing this test.

The question 1 says:

We say that a polynom is positivist if it can be wroted as a product of two non-constant polynoms with real coefficients $\geq 0$. Let be $f(x)$ a non-null polynom with constant coefficient non-null such that $f(x^n)$ is positivist for some positive integer $n$. Prove that $f(x)$ is positivist.

Well, let be $f(x^n)=p(x)q(x)$, where $p(x),q(x)$ are non-constant polynoms with real coefficients $\geq 0$.

The coefficients of $x^k$ with $n\not\mid k$ em $p(x),q(x)$ are nulls. This because the constant coefficients are $>0$, so we would have $x^k$ in the expression of $f(x^n)$, contradiction.

So, $p(x),q(x)$ can be wrote as $p(x^n),q(x^n)$.

We can done the variable changing $y=x^n$ and obtain $f(y)=p(y)q(y)$, where $p(y),q(y)$ are non-constant polynoms with real coefficients $\geq 0$, $QED$.

I don't know if this is correct and I'd like to have more opinions.

Quiet_waters
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    "This because the constant coefficients are $>0$" - I agree, but I'd elaborate more about why is that. $p(x),,q(x)$ can be written in the form $P(x^n),,Q(x^n)$ where $P,,Q$ are non-constant, we can't use the same notation here. I'd argue why $f(y)=P(y)Q(y)$ holds for all $y\in\mathbb{R}$, even for even $n$. If you make corrections, proof will be correct. – Jakobian Aug 21 '18 at 22:20
  • @Rumpelstiltskin, thanks very much. I'll elaborate more about the first citation and don't write with the same notation there. The last argument I'm trying yet, but I understand the question, thanks again. – Quiet_waters Aug 23 '18 at 00:34
  • @Rumpelstiltskin, I'm not getting the way to prove that $f(y)=P(y)Q(y)$ holds for $y<0$ to even $n$, since I cannot take $x=\sqrt[n]{y}$. Could you or somebody else help me? Thanks very much, – Quiet_waters Aug 23 '18 at 21:35
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    $f(y)-P(y)Q(y)$ is a polynomial with an infinite amount of zero's. From fundamental theorem of algebra, this can happen only when it's the $0$ polynomial, i. e. $f(y)-P(y)Q(y)=0$ for all $y\in \mathbb{R}$. I leave you to fill in the details. – Jakobian Aug 23 '18 at 22:39
  • @Rumpelstiltskin, I got understand (I'm wonderful with the idea). Thanks for the rich help. – Quiet_waters Aug 24 '18 at 00:14
  • Someone could help me to flag this topic as answered? Thanks and I'm sorry for any inconvenience. – Quiet_waters Aug 27 '18 at 21:35
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    I don't think anyone would mind if you just leave the topic, or delete it. But it'd be better to just leave it, maybe someone would have the same problem in the future – Jakobian Aug 28 '18 at 11:45
  • @Rumpelstiltskin, right, thanks very much again. – Quiet_waters Aug 28 '18 at 15:16

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