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Let $f(x)$ be a polynomial with degree $> 1$. If $f(x^n)$ can be decomposed as $q_1(x)q_2(x)$, $q_1$ and $q_2$ non-constant polynomials with non-negative coefficients, then prove that there exists $p_1(x)$ e $p_2(x)$ non-constant polynomials with non-negative coefficients such that $f(x) = p_1(x)p_2(x)$.

What I have done so far: If $f(x) = (x - \lambda_1) \dots (x - \lambda_m)$, then $ f(x^n) = (x^n - \lambda_1) \dots (x^n - \lambda_m)$

Thus $f(x^n) = (x - \lambda_1^{\frac{1}{n}})(x^{n-1} + \dots + (\lambda_1^{\frac{1}{n}})^{n-1}) \dots (x - \lambda_m^{\frac{1}{n}})(x^{n-1} + \dots + (\lambda_m^{\frac{1}{n}})^{n-1})$

We have that $q_1$ and $q_2$ can only have non-positive roots, so we can conclude that every $\lambda_i$, if real, is non-positive (otherwise $\lambda_i^{\frac{1}{n}}$ would be a positive root of $q_1q_2$). And that's all I have thought of. If every $\lambda_i$ were a non-positive real, we would be done (we could take $p_1(x) = (x + |\lambda_1| ) \dots (x + |\lambda_{m-1} |)$ and $p_2(x) = (x + |\lambda_m |)$). But I don't know what to do with the complex roots of $f$. I ask for hints, please.

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