Find all polynomials $p(x)$ satisfying the condition: $$p(x^2-2x)=p(x-2)^2.$$
Problem
Source: Bulgaria 1976 P2
Tags: fe, functional equation, algebra, Polynomials
Sprites
19.06.2021 09:29
Write $P(x)={a_1}x^n...+{a_n}$ Now $P(x^2-2x)={a_1}(x^2-2x)^n+...{a_n}$ And $P(x-2)^2={a_1}(x^2-4x+4)^n...+{a_n}$ Now $P(x-2)^2-P(x^2-2x)=0$ So ${a_1}(x^2-4x+4)^n...-{a_1}(x^2-2x)^n+...=0$ So it will only work if $P(x)=C$ where $C$ is a constant .(since the above implies that $P(x-2)$|$P(x)$)
rama1728
06.08.2021 11:02
Let \(z\) be a complex root of \(p(x)\). Then, substituting \(x=z+2\) in the given relation yields that \(z^2+2z=(z+1)^2-1\) is also a root. Reiterating the same thing, one can easily note that \((z+1)^{2^n}-1\) is a root for all \(n\geq 0\). Let the degree of \(p\) be \(d\), then \(p\) has at most \(d\) distinct roots. Then by PHP, since there can be at most \(d\) distinct roots, i.e \(d\) holes and the number of pigeons is the number of distinct \((z+1)^{2^n}-1\), we can say that there is some \(m\) for which \((z+1)^{2^m}-1=(z+1)^{2^{m+1}}-1\), implying that \((z+1)^{2^m}\left((z+1)^{2^m}-1\right)=0\), meaning that one of the roots of \(p\) should be \(0\) or \(-1\). If it is \(-1\), then substituting \(p(x)=(x+1)q(x)\) where \(q\) is another polynomial would give us the same polynomial but \(q\) in place of \(p\). Then we can make the same argument to deduce that \(p(x)=(x+1)^d\). Now, if that root is \(0\), then substituting \(p(x)=xq(x)\) where \(q\) is a polynomial would give \[xq(x^2-2x)=(x-2)q(x-2)\]Substituting \(2\) here in the new equation would give \(q(0)=0\), so \(0\) is a root of \(q\) also. Then we could define \(q(x)=xr(x)\) where \(r\) is another polynomial and analogously get \(0\) is a root of \(r\) and keep going, but note that \(p\) has only finitely many roots, so we will get that \(p(x)=x^d\) which is clearly not a solution. So if that root is \(0\), we have no solutions. Finally, if \(p\) is a constant polynomial, then we get that \(p(x)=0\) or \(p(x)=1\). Thus, the only solutions are \[p(x)=(x+1)^n,~p(x)=0,~p(x)=1\]
rama1728
06.08.2021 11:03
Sprites wrote: Write $P(x)={a_1}x^n...+{a_n}$ Now $P(x^2-2x)={a_1}(x^2-2x)^n+...{a_n}$ And $P(x-2)^2={a_1}(x^2-4x+4)^n...+{a_n}$ Now $P(x-2)^2-P(x^2-2x)=0$ So ${a_1}(x^2-4x+4)^n...-{a_1}(x^2-2x)^n+...=0$ So it will only work if $P(x)=C$ where $C$ is a constant .(since the above implies that $P(x-2)$|$P(x)$) \(p(x)=(x+1)^n\) is clearly a visible solution.
CROWmatician
11.08.2021 08:52
rama1728 wrote: Let \(z\) be a complex root of \(p(x)\). How do you know there exists a complex root? rama1728 wrote: Reiterating the same thing, one can easily note that \((z+1)^{2^n}-1\) is a root for all \(n\geq 0\). And how do we show it?
jasperE3
11.08.2021 08:56
CROWmatician wrote: rama1728 wrote:
Let \(z\) be a complex root of \(p(x)\). Then, substituting \(x=z+2\) in the given relation yields that \(z^2+2z=(z+1)^2-1\) is also a root. Reiterating the same thing, one can easily note that \((z+1)^{2^n}-1\) is a root for all \(n\geq 0\). Let the degree of \(p\) be \(d\), then \(p\) has at most \(d\) distinct roots. Then by PHP, since there can be at most \(d\) distinct roots, i.e \(d\) holes and the number of pigeons is the number of distinct \((z+1)^{2^n}-1\), we can say that there is some \(m\) for which \((z+1)^{2^m}-1=(z+1)^{2^{m+1}}-1\), implying that \((z+1)^{2^m}\left((z+1)^{2^m}-1\right)=0\), meaning that one of the roots of \(p\) should be \(0\) or \(-1\). If it is \(-1\), then substituting \(p(x)=(x+1)q(x)\) where \(q\) is another polynomial would give us the same polynomial but \(q\) in place of \(p\). Then we can make the same argument to deduce that \(p(x)=(x+1)^d\). Now, if that root is \(0\), then substituting \(p(x)=xq(x)\) where \(q\) is a polynomial would give \[xq(x^2-2x)=(x-2)q(x-2)\]Substituting \(2\) here in the new equation would give \(q(0)=0\), so \(0\) is a root of \(q\) also. Then we could define \(q(x)=xr(x)\) where \(r\) is another polynomial and analogously get \(0\) is a root of \(r\) and keep going, but note that \(p\) has only finitely many roots, so we will get that \(p(x)=x^d\) which is clearly not a solution. So if that root is \(0\), we have no solutions. Finally, if \(p\) is a constant polynomial, then we get that \(p(x)=0\) or \(p(x)=1\). Thus, the only solutions are \[p(x)=(x+1)^n,~p(x)=0,~p(x)=1\]
How do you know there exists a complex root? Otherwise, $f$ has degree $0$ and $f\not\equiv0$.
CROWmatician
11.08.2021 09:03
@above thanks, but i still don't fully get it, probably because my polynomials knowledge is weak, any handout/book suggestions for beginners?
pco
11.08.2021 09:03
CROWmatician wrote: How do you know there exists a complex root? If your question is "complex versus real", the words "complex root" here mean "any root in $\mathbb C$" (including real roots if any) If your question is "How do you know there exists a root (no matter it belongs to $\mathbb R$ or to $\mathbb C\setminus R$)?", this is a quite standard theorem that any polynomial in $\mathbb C[X]$ with degree $n\ge 1$ has exactly (including multiplicity) $n$ roots in $\mathbb C$.
a22886
11.08.2021 09:03
CROWmatician wrote: rama1728 wrote: Let \(z\) be a complex root of \(p(x)\). How do you know there exists a complex root? By the Basic Theorem for Algebra (I think this is the name ) there is always a complex root for any non-constant polynomial.
rama1728
11.08.2021 09:58
pco wrote: CROWmatician wrote: How do you know there exists a complex root? If your question is "complex versus real", the words "complex root" here mean "any root in $\mathbb C$" (including real roots if any) If your question is "How do you know there exists a root (no matter it belongs to $\mathbb R$ or to $\mathbb C\setminus R$)?", this is a quite standard theorem that any polynomial in $\mathbb C[X]$ with degree $n\ge 1$ has exactly (including multiplicity) $n$ roots in $\mathbb C$. Thank you very much for explaining this detailedly @pco!
jred
12.08.2021 11:01
a22886 wrote: CROWmatician wrote: rama1728 wrote: Let \(z\) be a complex root of \(p(x)\). How do you know there exists a complex root? By the Basic Theorem for Algebra (I think this is the name ) there is always a complex root for any non-constant polynomial. Well, we actually call it the fundamental theorem of algebra.
CROWmatician
12.08.2021 18:13
CROWmatician wrote: any handout/book suggestions for beginners? sooo, anything?
Bluecloud123
01.07.2024 10:46
CROWmatician wrote: CROWmatician wrote: any handout/book suggestions for beginners? sooo, anything? i assume you already found a great book since u got silver in imo so any recommendations for beginners like me?
MajesticCheese
02.07.2024 05:02
Let $q(x) = p(x-1)$
Then by the given assertion,
$q((x-1)^2) = q(x-1)^2$(The right hand side is p(x-2) and if you expand the (x-1)^2, you will get x^2 - 2x + 1 and then subtract that by 1)
Because this applies to all real $x$, just view this condition as $q(x^2) = q(x)^2$
Now, say $r_1, r_2 ... r_n$ are the roots of $q(x)$. Then the right hand side is $((x-r_1)...(x-r_n))^2$. Now the left side becomes $(x^2-r_1)...(x^2-r_n).$ (It is trivial that the leading coefficient of q(x) has to be 1)
Use difference of squares on the left hand side to get $(x-\sqrt{r_1})(x+\sqrt{r_1}).....(x-\sqrt{r_n})(x+\sqrt{r_n})$. Now comparing the roots of the left hand side and right hand side gives us that all the roots have to be equal to 0(or else infinite descent, or even setting it equal to 1 or -1 doesn't do the trick)
Hence $q(x) = x^n.$ Because $p(x-1) = q(x),$ $p(x) = (x+1)^n$
MajesticCheese
02.07.2024 05:05
Bluecloud123 wrote: CROWmatician wrote: CROWmatician wrote: any handout/book suggestions for beginners? sooo, anything? i assume you already found a great book since u got silver in imo so any recommendations for beginners like me? I'm not an IMO qual : but here's a handout: https://alexanderrem.weebly.com/uploads/7/2/5/6/72566533/polynomials.pdf