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

Monday, July 12, 2021

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

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

(2010 Baltic Way)

Noted that: $2010=30.67$

We can write that:

$P(x)=(x-67)Q(x)$

We can write that:

$P(x)=(x-2010)G(x)$

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, $P(x)=(x-67)(x-2.67)(x-3.67)........(x-30.67)C |C\inR$

Math Book Cambodia