# Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$

Find all polynomials $P$ such that

$P(x^2+1)=P(x)^2+1$

-
Is $P$ a real polynomial? –  Frank Science Jan 6 '13 at 6:12
$P(x)=x$ works. Don't know if there are others. Nice question. –  coffeemath Jan 6 '13 at 6:53
@Coffeemath Let $\alpha$ satisfy $\alpha=\alpha^2+1$ then $p(x)=\alpha$ for all $x$ works as well. –  Amr Jan 6 '13 at 7:14
I found the problem here: imomath.com/index.php?options=346&lmm=0 The solution is the same as miracle173's. –  sdcvvc Jan 7 '13 at 13:56

One solution is $P(x) = x^2 + 1$:

$$P(x^2 + 1) = (x^2 + 1)^2 + 1 = P(x)^2 + 1$$

Another solution is $P(x) = x^4 + 2x^2 + 2$:

\begin{align} P(x^2 + 1) &= (x^2 + 1)^4 + 2(x^2 + 1)^2 + 2 \\&= x^8 + 4 x^6 + 8 x^4 + 8 x^2 + 5 \\&= (x^4 + 2x^2 + 2)^2 + 1 \\&= P(x)^2 + 1 \end{align}

How did I find these? Turns out it's obvious after rewriting the equation! Let $Q(x) = x^2+1$. Then the equation is

$$P(Q(x)) = Q(P(x))$$

or more succinctly, $P \circ Q = Q \circ P$. It's now clear that $P=Q$ is one solution, $P = Q \circ Q$ is another solution, $P = Q \circ Q \circ Q$ is yet another, and so forth.

Note that $P(x) = x$ is in this family as well, as the identity function is the empty repeated composition (just like the empty sum is $0$ and the empty product is $1$).

This argument is easily adapted to show the set of solutions is a monoid under composition.

I haven't worked out the complete solution though. Alas I don't know much about the monoid of polynomials under composition.

This answer to a similar question cites a 1922 theorem of Ritt that contains a characterization of polynomials that commute under composition; in particular, we can conclude the repeated compositions of $Q$ are indeed the entire solution space, excluding nonconstant polynomials. The remaining solutions are thus the two functions $P(x) = \beta$ where $\beta$ is a solution to $\beta = Q(\beta)$.

-
Aha. Nice observation. –  Amr Jan 6 '13 at 7:47
Interesting that you did not start with $P(x) = x$. –  user17762 Jan 6 '13 at 8:37
@Marvis There is also another solution, Let α satisfy $α=α^2+1$ then $p(x)=α$ for all x works as well. –  Amr Jan 6 '13 at 8:47
I suspect that all non-constant solutions are compositional powers of $x^2+1$. I also suspect that asc.tuwien.ac.at/funkana/woracek/papers/ppol.pdf is relevant here. –  Robert Israel Jan 6 '13 at 10:35

A full characterization of solutions which is quite short and sweet.

Plugging in $x = \pm a$, we see that $P(x) = \pm P(-x)$ for each $x$. Clearly one sign holds infinitely often. By looking at the roots of the polynomial $P(x) - P(-x)$ or $P(x) + P(-x)$, we easily see that $P(x)$ is either odd or even. If $P(x)$ is odd, then note that $P(0) = 0$. Then $P(1) = 1$, $P(2) = 2$, $P(5) = 5$, etc. and by induction we can get infinitely many values $a$ such that $P(a) = a$. By looking at the roots of the polynomial $P(x) - x$ we see that $P(x) = x$ for all $x$ then.

Now suppose $P$ is even. This means $P(x) = Q(x^2)$ for some polynomial $Q$. Let $R(x) = x^2 + 1$. Remark there clearly exists a polynomial $S(x)$ such that $P(x) = S(x^2+1)$, just shift $Q$. Now remark that $P \circ R = R \circ P$. Hence $S \circ R \circ R = R \circ S \circ R \implies S \circ R = R \circ S$, so $S$ is a solution to the functional equation. By an easy induction on degree, it follows the only even solutions to the equations are iterations of $R$, so we are done. (EDIT : Also there is the constant solution $P(x) = c$ where $c = c^2 + 1$, because the induction starts on degree 1 which I forgot)

-
(two constant solutions) –  Hurkyl Jan 8 '13 at 7:30
the first part i exactly the same as in my solution. The second parts avoids using the sqareroot. that is nice. –  miracle173 Jan 8 '13 at 7:45

There is another method (besides that posted here) to prove that solutions have the structure $\sum_{k=0}^n\,a_{2k}x^{2k}$ without constructing formulas to

calculate the coefficients as in my other answer.

We have $$P(x)^2-P(-x)^2=(P(x^2+1)-1)-(P((-x)^2+1)-1)=0$$

At least one of $P(x)-P(-x)$ and $P(x)+P(-x)$ must have infinitely many zeros and therefore must be identical to $0$.

Asume that $P(-x)=-P(x)$. This means that $P$ is an odd function and $P(0)=0$. Let us define $$Q(x)=x^2+1$$ then $P(Q^n(0)=Q^n(0)$ and so $$P(x)-x=0, \quad x=0,\,Q(0),\,Q^2(0),\ldots$$ for infintely many $x$, so $P(x)=x$.

$P$ must be an even polynomial if not $P(x)=x$. But the even polynomials are exactly the polynomial that contain only even powers of x.

The polynomials $Q^n$ are even polynomials that satisfy the functional equation. They have a degree of $2^n$

In contrast to my other answer I was not able to prove with this method that there is at most one polynomial of a certain degree that satisfy the functional equation.

Proof
a polynomial is even $\Longleftrightarrow$ the polynomial contains only even powers of $x$

The even polynomial $P(x)$ can be expressed as sum $f(x)+g(x)$ were $f(x)$ is a polynomial that only contains even powers of $x$ and $g(x)$ is a polynomial that contains only odd powers of $x$. We have $$P(x)-f(x)=g(x)$$ The left side is an even polynomial, the right side an odd one, so $P(x)=f(x)$. So polynomials that contain only even powers of $x$ are exactly the even polynomials.

EDIT:

The following completes the proofs:

Lemma $Q(x)=x^2+1$,$P(Q(x)=Q(P(x))$ and $P(x)$ is even. Then there is a polynomial $T(x)$ with $P=T \circ Q = Q \circ T$

So a polynomial $P$ of degree $n$ can be reduced to a polynomial with degree $\frac{n}{2}$ that also satisfies the functional equation. This process can repeated until we

arrive at a polynomial with odd degree. But the only polynomial with odd degree that satisfies the functional equation is $T(x)=x$.

Therefore the functional equation $$Q \circ P = P \circ Q$$ has exactly the following polynomial solutions:

$$-\frac{\sqrt{3}i-1}{2} \\ \frac{\sqrt{3}i+1}{2} \\ x \\ Q(x)\\ Q^2(x) \\ Q^3(x) \\ \ldots$$

Proof of the Lemma
We define $R(x)=\sqrt[+]{x-1}$, then $$R(Q(x))=Q(R(x))=x,\quad \forall x>1$$ and $$R(x)^2=x-1, \quad \forall x>1$$ We have $$Q \circ P = P \circ Q$$ and therefore $$Q \circ P \circ R = P \circ Q \circ R = P = P \circ R \circ Q , \quad \forall x>1$$ For $x> 1$ the polynomial $$T(x)=P(R(x))=\sum_{k=0}^{n}a_{2n}(x-1)^{n}$$ satisfies $$P=T \circ Q = Q \circ T$$ But if a polynomial equation is satisfied for infinite many $x$ it is satisfied for all $x$.

The coefficients of the polynomials can be efficiently calculated by the formulae $(3)$ and $(4)$ of my other post

-
Are you sure you meant that degree of $P$ is reduced from $n$ to $n-2$, not $n/2$? –  sdcvvc Jan 7 '13 at 14:48
@ sdcvvc: thank you, fixed. –  miracle173 Jan 7 '13 at 15:03
Ah! I had thought to look at the inverse of $Q$ and its iterates, but it hadn't clicked that $P \circ Q^{-1}$ is a polynomial! Well done. –  Hurkyl Jan 8 '13 at 6:23

Let $P(y)=\sum_{0\le r\le n}a_ry^r$

So, $$P(1+x^2)=\sum_{0\le r\le n}a_r(1+x^2)^r=a_0+a_1(1+x^2)+a_2(1+\binom 21x^2+x^4)+\cdots +a_{n-1}(1+\binom {n-1}1x^2+\binom {n-1}2x^4+\cdots+\binom {n-1}{n-2}x^{2(n-2)}+\binom {n-1}{n-1}x^{2(n-1)}) +a_n(1+\binom n1x^2+\binom n2x^4+\cdots+\binom n{n-1}x^{2(n-1)}+\binom n nx^{2n})$$

$$=x^{2n}a_n+x^{2n-2}(a_n\binom n{n-1}+a_{n-1})+x^{2n-4}(a_n\binom n{n-2}+a_{n-1}\binom {n-1}{n-2}+a_{n-2})+x^2(a_n\binom n1+a_{n-1}\binom{n-1}1+\cdots+a_2\binom21+a_1)+\sum_{0\le r\le n}a_r$$

and $$\{P(x)\}^2+1=\{\sum_{0\le r\le n}a_rx^r\}^2+1$$ $$=a_n^2x^{2n}+x^{2n-1}2a_na_{n-1} +x^{2n-2}(a_{n-1}^2+2a_na_{n-2})+x^{2n-3}2(a_na_{n-3}+a_{n-1}a_{n-2}) +x^{2n-4}(a_{n-2}^2+2a_na_{n-4}+2a_{n-1}a_{n-3})+\cdots+x^2(a_1^2+2a_0a_2)+\sum_{0\le r\le n}a_r^2+1$$

Comparing the coefficients of the different powers of $x$

$r=n\implies a_n=a_n^2\implies a_n=1$ as $a_n\ne0$

$r=n-1\implies 2a_na_{n-1}=0\implies a_{n-1}=0$

$r=n-2\implies a_n\binom n{n-1}+a_{n-1}=a_{n-1}^2+2a_na_{n-2}\implies a_{n-2}=\frac n2$

$r=n-3\implies 2(a_na_{n-3}+a_{n-1}a_{n-2})=0\implies a_{n-3}=0$

$r=n-4\implies a_n\binom n{n-2}+a_{n-1}\binom {n-1}{n-2}+a_{n-2}=a_{n-2}^2+2a_na_{n-4}+2a_{n-1}a_{n-3}\implies a_{n-4}=\frac {n^2}8=\frac1{2!}\left(\frac n2\right)^2$

$r=n-5\implies 2(a_na_{n-5}+a_{n-1}a_{n-4}+a_{n-2}a_{n-3})=0\implies a_{n-5}=0$

$r=n-6\implies 2(a_na_{n-6}+a_{n-1}a_{n-5}+a_{n-2}a_{n-4})+a_{n-3}^2=a_n\binom n{n-3}+a_{n-1}\binom{n-1}{n-3}+a_{n-2}\binom{n-2}{n-3}+a_{n-3}\implies a_{n-6}=\frac1{3!}\left(\frac n2\right)^3-\frac n3$

-
What about the rest of the cofficents ? –  Amr Jan 6 '13 at 8:19
@Amr, I was just trying to find the pattern, observe that $n=2$ gives us Hurkyl's solution. –  lab bhattacharjee Jan 6 '13 at 8:36

From @lab bhattacharjee post we see that the formulae for $P(1+x^2)$ and $P(x)^2+1$ are

$$\begin{eqnarray} P(1+x^2)&=&\sum_{k=0}^{n}{\left(\sum_{i=0}^{n-k}{a_{n-i}\,{{n-i}\choose{k}}} \right)\,x^{2\,k}} \tag{1} \\ P(x)^2+1&=&\sum_{k=0}^{n}{\left(\sum_{i=0}^{k}{a_{n-i}\,a_{n-k+i}}\right)\,x^{ 2\,n-k}} +\sum_{k=0}^{n-1}{\left(\sum_{i=0}^{k}{a_{i}\,a_{k-i}} \right)\,x^{k}}+1 \tag{2} \\ \end{eqnarray}$$

The coefficients of the monomials with odd exponents in $(1)$ are $0$. For the coefficients of the exponent $2n$ we get the equation

$$a_{n}^2=a_{n}$$ for $n>0$ and therefore $$a_{n}=1$$ If $a_{n}=0$ then it would be $degree(g)<n$. For $n=0$ one gets the equation $a_{0}^2+1=a_{0}$ instead.

The remaining $a_i$ can calculated successively. After calculation $a_n \ldots a_{n-(k-1)}$ for even $k=2m$ the value $a_{n-k}$ can be calculated by the coefficient

of $x^{2n-2m}$

$$\sum_{i=0}^{2m}{a_{n-i}\,a_{n-2m+i}}= \sum_{i=0}^{m}{a_{n-i}\,{{n-i}\choose{n-m}}}$$

which gives

$$a_{n-2m}=\frac{\sum_{i=0}^{m}{a_{n-i}\,{{n-i}\choose{n-m}}}-\sum_{i=1}^{2m-1}{a_{n-i}\,a_{n-2m+i}}}{2a_n} \tag{3}$$

indexes of the $a$ on the right hand side of this equation are all larger than $2n-2m$ and therefore already known. For odd $k=2m+1$ we got the equation

$$\sum_{i=0}^{2m+1}{a_{n-i}\,a_{n-2m-1+i}}=0$$

and therefore

$$a_{n-2m-1}=-\frac{\sum_{i=1}^{2m}{a_{n-i}\,a_{n-2m-1+i}}}{2a_n} \tag{4}$$

This shows that $a_{n-2m-1}$ must be $0$. This can be proven by inductions starting with $m=0$ whch gives $a_{n-1}=0$ from $(4)$. In equation $(4)$ one factor $a_j$ of

each summand has an odd index because 2m-1 is odd and so $a_i=0$ and also the whole summand by induction. So the whole sum is $0$. So $0=a_{n-1}=a_{n-3}=...$. For $n$ odd this means especially $a_0=0$.

With the formulae $(3)$ and $(4)$ we can calculate the sequence $a_n, a_{n-1}, \ldots, a_0$ from the coefficients of $x^{2n}, \ldots, x^{n}$. But it is possible that

$a_i$ calculated do not fullfil the equations for the coefficients of $x^{0}, \ldots, x^{n-1}$. These coefficients give raise to the following equations

$$\sum_{i=0}^{2m}{a_{i}\,a_{2m-i}}=\sum_{i=0}^{n-m}{a_{n-i}\,{{n-i}\choose{m}}} \tag{5}$$

or $$a_0^2+1=\sum_{i=0}^{n} a_i$$

if $k=m=0$

$$\sum_{i=0}^{2m+1}{a_{i}\,a_{2m+1-i}}=0 \tag{6}$$

So we can only condlude that there is at most one polynomial for each degree. The polynomials from the the post of @Hurkyl shows that there exists polynomial with $degree(P)=2^n$ for each $n$.

From @Hurkyl We take the definition of the polynomial

$$Q(x)=x^2+1$$

and we know that the problem can be restated as

Find all polynomial $P$ that $$P(Q(x))=Q(P(x))$$ or $$P \circ Q = Q \circ P$$

therefore

$$P \circ Q^n = Q^n \circ P$$

and @Hurkyl showed that all

$$P=Q^n$$ are solutions

I checked for $degree(P)$ from $0$ to $16$ that there are only the following polynomials:

$$-\frac{\sqrt{3}i-1}{2} \\ \frac{\sqrt{3}i+1}{2} \\ x \\ x^2+1 \\ x^4+2x^2+2 \\ x^8+4x^6+8x^4+8x^2+5 \\ x^{16}+8x^{14}+32x^{12}+80x^{10}+138x^8+168x^6+144x^4+80x^2+26 \\$$

Besides the constatn solutions the only solutions that exists where the solutions @Hurkyl found.

Let $P$ be a polynomial that fullfills the equation $P * Q = Q *P$. We showed that $a_0$=0 if $degree(P)$ is odd. Therefore $P(0) =0$ and also

$$P(Q^n(0))=Q^n(P(0))=Q^n(0)$$

So $Q^n(0), n=0,1,2,3,\ldots$ is a strictly increasing and therefore infinite sequence with $P(x)=x$ and therefore $P(x)-x=0$. But if a polynomial $P(x)-x$ is $0$ for

infinite many values $x$ the the polynomial es equal to $0$. So $P=id$. THis means that $x$ is the only polynomial with odd degree that satisfies our functional equation.

The remaining problem: Show that $degree(P)=2^n$ if $degree(P)$ is even.

Edit:
The solution fo the remaining problem is now already included in my other post

-
addeda proof that there are no solutions with odd degree –  miracle173 Jan 7 '13 at 0:50

It depends on whether the coefficients of the polynomial come from a field (or ring) of characteristic $0$. [ http://en.wikipedia.org/wiki/Characteristic_%28algebra%29 ]

There are extra solutions in characteristic $p$, such as $P(x) = x^p$. It could be a hard problem to determine whether solutions exist that are not generated by compositions of $x^p$ and $x^2+1$. The general problem of determining all commuting pairs of polynomials over a finite field is well-known and unsolved.

In characteristic $0$, solutions of $P(Q(x))=Q(P(x))$ are classified. The case $Q(x)=x^2+1$ does not fall into one of the families of nontrivial solutions (it is not linearly conjugate to $x^2$ or to the degree 2 Chebyshev polynomial), so that the only possibililty is $P(x) = Q^{\circ \hskip0.7pt n}$ for some $n$.

-