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Does there any theorem with algorithm which says, any polynomial $P$ and $Q$ with common variable then $P$ can be represented in terms of $Q$ as $$P=\sum_{i\in S}P_iQ^i$$ and there exists unique polynomials $P_i$ have same degree for all $i$ and $S\subset\mathbb{Z}$.

For example: Sum of cube of first natural number is square of sum of first natural number that is

$P(n) = \sum_{k=1}^n k^3$ and $Q(n)= \sum_{k=1}^n k$ and $P(n)= 1\cdot Q^2(n)+0\cdot Q(n)+\ldots$

I'm not taking about polynomial division algorithm. Thanks.

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  • $\begingroup$ I don't think $P=x$ can be represented in terms of powers of $Q=x^2$... Or do you mean $P_0 = x^2$? Question is somewhat unclear. $\endgroup$ Commented Apr 8, 2023 at 0:06
  • $\begingroup$ Do you allow the case where $\,S=\{0\}$? $\endgroup$
    – Somos
    Commented Apr 8, 2023 at 0:10
  • $\begingroup$ @Somos yes, if $deg(Q)>deg(P)$ then $S=\{0\}$ only option will be there $\endgroup$
    – Pruthviraj
    Commented Apr 8, 2023 at 1:20
  • $\begingroup$ @OleksandrKulkov if $P=x$ and $Q=x^2$ then $P=P_0$ because $deg(Q)>deg(P)$ $\endgroup$
    – Pruthviraj
    Commented Apr 8, 2023 at 1:34

1 Answer 1

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If you specifically want that $\deg P_i < \deg Q$, the existence of such representation indeed follows from the fact that polynomials over any field form a Euclidean domain, that is, we can uniquely represent

$$ P(x) = P_0(x) + D_1(x) Q(x), $$

where $\deg P_0 < \deg Q$. Then, we can in turn represent

$$ D_1(x) = P_1(x) + D_2(x) Q(x), $$

where $\deg P_1 < \deg Q$, and so on, so you can find $P_0, P_1, \dots$ with consecutive division by $Q$ and taking remainders. Note that, generally, $\deg D_k = \deg P - k \deg Q$, so it will get down to $D_k = 0$ eventually.

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  • $\begingroup$ Is this the only way (know so far) to represent a polynomial in terms of another polynomial $\endgroup$
    – Pruthviraj
    Commented Apr 8, 2023 at 16:47
  • $\begingroup$ Yes, it is unique, as remainders and long division result are unique. $\endgroup$ Commented Apr 8, 2023 at 20:23

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