abeker wrote: Determine all pairs of polynomials $(P, Q)$ with real coefficients satisfying $$P(x + Q(y)) = Q(x + P(y))$$for all real numbers $x$ and $y$. From $P(x + Q(y)) = Q(x + P(y))$ we get $Q(x)=P(x+(Q(y)-P(y)))(*)$. Case $1$. $P(x)-Q(x)$ is non-constant. If $Q(y)-P(y)$ is non-constant, then there exists infinitely many real numbers which can be written as $Q(y)-P(y)$ for some real $y$, so we have $P$ is constant since $P(x+(Q(a)-P(a))=P(x+(Q(b)-P(b)))=Q(x)$. $p \equiv c ,(*) \implies Q$ is constant $\implies P(x)-Q(x)$ is constant; contradiction. Case$2$. $P(x)-Q(x)$ is constant. So we can assume $Q(x)-P(x)=t(**)$ for some real $t$. $(*),(**) \implies P(x+t)=P(x)+t$, this is an easy equation.(if $t \not = 0$ prove that $P(x)=x+a$) So $\boxed {P=Q}$ or $\boxed {P(x)=x+a,Q(x)=x+b}$ for some real $a,b$.

This problem was proposed by me. Another approach is to write $P(x)=ax^n+bx^{n-1}+R(x)$ and $Q(x)=cx^m+dx^{m-1}+S(x)$ with $n,m\ge 1$, $a\neq 0 \neq c$, $\deg R < n-1$, $\deg S < m-1$. Comparing degrees and the leading coefficients we find $m=n$ and $a=c$. Comparing coefficient of $x^{n-1}$ we get $anQ(y)+c=anP(y)+d$ for every $y \in \mathbb R$. It follows that $P-Q$ is constant. If $P-Q$ is zero then $P=Q$ which clearly works. Otherwise we get that $Q(x+t)-Q(x)$ (for some fixed number $t\neq 0$) is constant which easily implies that $Q$ is linear so $P$ is linear so we only have to check linear functions which leads to $P(x)=x+a$ and $Q(x)=x+b$.

(1) By setting $y=0$ and $x=z-P(0)$ and by defining constant $c=Q(0)-P(0)$, we get $Q(z)=P(z+c)$ for all $z$. Hence $Q$ is just a shift of $P$. (2) If $c=0$, we get the solutions with $P=Q$. Hence assume from now on that $c\ne0$. The equation turns into $P(x+P(y+c))=P(x+P(y)+c))$ (3) If there exists a real number $r$ with $P(r+c)\ne P(r)+c$, then we set $x=z-P(r+c)$ and $y=r$ and $d=P(r)+c-P(r+c)\ne0$ to get $P(z)=P(z+d)$ for all $z$. As the only periodic polynomials are the constant ones, we get the solution $P(x)=Q(x)=const$. (4) In the only remaining case we have $P(r+c)=P(r)+c$ for all real $r$. By induction we prove $P(nc)=nc+ P(0)$ for all positive integers $n$. Hence there are infinitely many real numbers $x$ for which $P(x)=x+P(0)$. By the identity theorem for polynomials, this implies that $P(x)=x+a$ is linear and that $Q(x)=P(x+c)=x+b$ is linear. This forms the last solution.

m.yetka wrong solution

m.yekta wrote: abeker wrote: Determine all pairs of polynomials $(P, Q)$ with real coefficients satisfying $$P(x + Q(y)) = Q(x + P(y))$$for all real numbers $x$ and $y$. From $P(x + Q(y)) = Q(x + P(y))$ we get $Q(x)=P(x+(Q(y)-P(y)))(*)$. Case $1$. $P(x)-Q(x)$ is non-constant. If $Q(y)-P(y)$ is non-constant, then there exists infinitely many real numbers which can be written as $Q(y)-P(y)$ for some real $y$, so we have $P$ is constant since $P(x+(Q(a)-P(a))=P(x+(Q(b)-P(b)))=Q(x)$. $p \equiv c ,(*) \implies Q$ is constant $\implies P(x)-Q(x)$ is constant; contradiction. Case$2$. $P(x)-Q(x)$ is constant. So we can assume $Q(x)-P(x)=t(**)$ for some real $t$. $(*),(**) \implies P(x+t)=P(x)+t$, this is an easy equation.(if $t \not = 0$ prove that $P(x)=x+a$) So $\boxed {P=Q}$ or $\boxed {P(x)=x+a,Q(x)=x+b}$ for some real $a,b$. $p (a) -q (a) $can be finite and have different values, which you have not seen.

but $P(x)-Q(x)$ is as well a polynomial, if it is finite and has different values at different points, then it is not constant, so it has some degree $\implies$ it can have infinite points

Lemma:$a \leq a_n x^n+...+a_1x+a_0 \leq b$ if this condition is satisfied, $n = 0$.