Let $P\_1,\dots,P\_k$ be polynomials over $\mathbf{C}$, no two of them being proportional.


Does there exist an integer $N$ such that $P\_1^N,\dots,P\_k^N$ are linearly independent?



Are large powers of polynomials linearly independent?
The answer is yes. In fact, an even stronger claim is true: there exists some $N$ such that for all $n \geq N, \ P\_{1}^{n}, \dots, P\_{k}^n$ are linearly independent over $\mathbb{C}$.


For this we will use a generalization of the [Mason-Stother's theorem](https://en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem) which appears on the Wikipedia page (though I have taken the special case of the curve $C = \mathbb{P}^{1} (\mathbb{C})$ and written it in slightly different language.):


Let $q\_1, \dots, q\_{k}$ be polynomials such that $q\_1 + \cdots + q\_{k} = 0$ and every proper subset of $q\_1, \dots, q\_{k}$ is linearly independent. Then,
$$\max \left\{ \mathrm{deg} \left( q\_1 \right), \dots, \mathrm{deg} \left( q\_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q\_1 \cdots q\_{k} \right) \right) - 1\right)$$




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Now, we can prove the claim by induction on $k$. For $k = 2$ it is obvious. Now, by induction for all $n$ large enough every proper subset of $P\_{1}^{n}, \dots, P\_{k}^{n}$ is linearly independent. Suppose for contradiction that there exist constants $\lambda\_{1}, \dots, \lambda\_{k}$ such that
$$\lambda\_1 P\_{1}^n + \cdots + \lambda\_{k} P\_{k}^n = 0$$
Letting $q\_i = \lambda\_i P\_{i}^{n}$, notice that $q\_1, \dots, q\_k$ satisfy the requirements of the lemma (we have assumed that $\lambda\_i \neq 0$), and therefore
$$n \leq \max \left\{ \mathrm{deg} \left( q\_1 \right), \dots, \mathrm{deg} \left( q\_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q\_1 \cdots q\_{k} \right) \right) - 1\right) = \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( P\_1 \cdots P\_k \right) \right) - 1 \right)$$
but the right hand side is constant, and so for $n$ large we get a contradiction.


