Find all Polynomials `P(x)` which is satisfied that: `(x-2010)P(x+67)=xP(x)` (2010 Baltic Way)

 Find all Polynomials `P(x)` which is satisfied that:

`(x-2010)P(x+67)=xP(x)`

(2010 Baltic Way) 

Noted that: `2010=30.67`

• If `x=0\leftrightarrow(-2010)P(67)=0\rightarrowP(67)=0`

We can write that: `P(x)=(x-67)Q(x)`

• If `x=2010\leftrightarrow0P(2007)=2010P(2010)\rightarrowP(2010)=0`

We can write that: `P(x)=(x-2010)G(x)`

• If `x=67\leftrightarrow(67-2010)P(2.67)=67P(67)\rightarrowP(2.67)=0`
• If `x=2.67\leftrightarrow(2.67-2010)P(3.67)=2.67P(2.67)=0\rightarrowP(3.67)=0`

As the same replacing, we consider that for all `1\leqn\leq30` then `P(n.67)=0`

we will prove this statement by Mathematics Induction 

For all `k\inn` and `1\leqn\leq30` we will prove `P(k.67)=0`

`k=0\rightarrowP(0)=0` It is true for `k=0`

We are assuming that's true with `k=n\rightarrowP(n.67)=0``(*)`

We will prove It is true till `k=n+1\rightarrowP[(n+1)67]=0`?

From: `(x-2010)P(x+67)=xP(x)`

Then, `(n.67-2010)P(n.67+67)=n.67P(n.67)=0` Because from our assuming!`(*)`

            `\rightarrowP(n.67+67)=0\leftrightarrowP[(n+1).67]=0` 

Thus, `P(n.67)=0` for all `1\leqn\leq30`

We can write that:

`P(x)=(x-67)(x-2.67)(x-3.67)........(x-30.67)Q(x)`

Then,    `xP(x)=x(x-67)(x-2.67).....(x-30.67)Q(x)`    `(1)`

From our hypothesis: `(x-2010)P(x+67)=xP(x)`

But, `(x-2010)P(x+67)=(x-2010)x(x-67)(x-2.67).....(x-29.67)Q(x+67)`    `(2)`

From both equations `(1)` and `(2)`: we will see that 

                `Q(x)=Q(x+67)` 

We easily see that `Q(x)=C | C\inR`
Hence, `P(x)=(x-67)(x-2.67)(x-3.67)........(x-30.67)C |C\inR`