Solution January 31 2007

Problem

Find all real polynomials such that

$p(x+p(x))=p(x)+p(p(x))\qquad\qquad\qquad(1)$

Solution I by int_e

Note that if a polynomial equation (i.e. an equation with polynomials in $x\,$ on both sides) is true for infinitely many real numbers, it is satisfied for all complex numbers. To see why, consider the difference of the two sides as a polynomial in $x\,$ over the complex numbers. For real $x\,$ , the polynomial evaluates to the same value over the complex numbers as over the real numbers, so it has infinitely many zeros over the complex numbers. It follows that it is the zero polynomial, and the equation is satisfied for all complex numbers, as claimed. So we will solve the problem for complex polynomials first, and then restrict the solutions to the real polynomials.

We will consider a few cases.

If $p(x)\,$ is a constant polynomial, $p(x) = k\,$ , then $k = k+k\,$ , i.e. $k=0\,$ and $p(x) = 0\,$ . This is a solution of the problem.

Otherwise, if all zeros of $p(x)\,$ are $0\,$ , i.e. $p(x) = kx^l\,$ with $l>0\,$ and $k\neq 0\,$ , then

$k(x+kx^l)^l = k x^l + k^{l+1} x^{l^2}\,$

$(1+kx^{l-1})^l = 1 + (kx^{l-1})^l\,$

The left side has $l+1\,$ non-zero terms after expanding, while the right side has $2\,$ . By comparing coefficients we see that these two numbers must equal. It is easily verified that $p(x) = kx\,$ is indeed a solution of the problem for all $k\in\mathbb{C}\,$ .

In all other cases, $p(x)\,$ has a zero $p(a)=0\,$ with $a\neq 0\,$ , by the fundamental theorem of algebra. (http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra) Likewise, there's a root $b_0\,$ of $p(x)-a=0\,$ .

Now if $p(b)=a\,$ then by (1),

$p(b+a) = p(b + p(b)) = p(b) + p(p(b)) = a + 0 = a\,$

By induction it follows that $p(b_0 + an)=a\,$ for all natural numbers $n\,$ . Thus, $p(x)-a\,$ has infinitely many zeros and must be constant. This means that $p(x)\,$ is constant as well, contrary to our assumptions. $\square$

In summary, the solutions are the polynomials $p(x) = kx\,$ with $k\in\mathbb{R}$ . (Note that this includes the first case)

Solution II by landen

Find all real polynomials such that

$p(x+p(x))=p(x)+p(p(x))\qquad\qquad\qquad(1)$

By direct substitution we get that $p(x)=a_1\,x$ is a solution for any real $a_1\,.$

Let $p(x)=a_n\,x^n +a_{n-1}\,x^{n-1}+\cdots a_1\,x + a_0$ with $n>1\,$ and $a_n\neq 0\,.$

Letting $x=0,\,$ we have $p(p(0))=p(0)+p(p(0))\,$ and $p(0)=a_0=0.\,$

For simpler notation take $p(x) = a_n\,x^n + q_{n-1}\,(x)$ where $q_{n-1}(x)\,$ is a polynomial of lower degree than $n.\,$

By expansion of the left side of (1) we get.

$p(x+p(x))=p(x)+p(p(x)) +\mbox{ cross terms between }p\mbox{ and } x.\,$

The cross term of highest degree must have a least one $x\,$ to be a cross term and then as many powers of $a_n x^n\,$ as possible. Any powers of $q_{n-1}\,$ will give smaller cross terms.

$a_n\,(x+a_n\,x^n)^n=a_n\,n\,x\,(a_n\, x^n)^{n-1}+\dots=n\,a_n^n\,x^{n^2-n+1}+\dots$ using the binomial theorem.

$p(x+p(x))-(p(x)+p(p(x)))=0=n\,a_n^n\,x^{n^2-n+1}+$ lower order cross terms.

$a_n=0\,$ which violates our assumption that it was nonzero for some $n>1.\,$ Therefore, there are no solutions of degree $n>1.\,$

$\square$

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