(2010 Baltic Way)
Noted that: \(2010=30\times67\)
If \(x=0\), then:
\((-2010)P(67)=0\rightarrow P(67)=0\)
We can write that:
\(P(x)=(x-67)Q(x)\)
If \(x=2010\), then:
\(0P(2007)=2010P(2010)\rightarrow P(2010)=0\)
We can write that:
\(P(x)=(x-2010)G(x)\)
If \(x=67\), then:
\((67-2010)P(2\times67)=67P(67)\rightarrow P(2\times67)=0\)
If \(x=2\times67\), then:
\((2\times67-2010)P(3\times67)=2\times67P(2\times67)=0\rightarrow P(3\times67)=0\)
By the same process, we consider that for all \(1\leq n\leq30\), \(P(n\times67)=0\).
We prove this statement by mathematical induction.
For \(k=0\):
\(P(0)=0\), it is true.
Assume that for \(k=n\):
\(P(n\times67)=0\)
We prove that:
\(P((n+1)\times67)=0\)
From:
\((x-2010)P(x+67)=xP(x)\)
Let \(x=n\times67\), then:
\((n\times67-2010)P(n\times67+67)=n\times67P(n\times67)=0\)
Therefore:
\(P((n+1)\times67)=0\)
Thus, \(P(n\times67)=0\) for all \(1\leq n\leq30\).
Hence:
\(P(x)=(x-67)(x-2\times67)(x-3\times67)\cdots(x-30\times67)Q(x)\)
Then:
\(xP(x)=x(x-67)(x-2\times67)\cdots(x-30\times67)Q(x)\) (1)
From the hypothesis:
\((x-2010)P(x+67)=xP(x)\)
Also:
\((x-2010)P(x+67)= (x-2010)x(x-67)(x-2\times67)\cdots(x-29\times67)Q(x+67)\) (2)
Comparing (1) and (2), we get:
\(Q(x)=Q(x+67)\)
Therefore, \(Q(x)=C\), where \(C\in\mathbb{R}\).
Hence, all polynomials are:
\[ P(x)=C(x-67)(x-2\times67)(x-3\times67)\cdots(x-30\times67),\quad C\in\mathbb{R} \]
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