Denote $\deg P(x) = p$, and wlog $\deg Q(x) = q \geq r = \deg R(x)$. If $P$ is constant or $Q$ is constant or $R$ is identically null, the claim is trivial, so assume $p,q >0$, $r\geq 0$. Define $C_k (f(x))$ to be the coefficient of $x^k$ in the polynomial $f(x)$, so $C_{\deg f(x)} (f(x)) \neq 0$ is the leading coefficient of $f(x)$. Also make the notations $a = C_{\deg P(x)} (P(x))$, $b = C_{\deg Q(x)} (Q(x))$, $c = C_{\deg R(x)} (R(x))$. If $q>r$, then $C_{pq} (P(Q(x)) + P(R(x))) = ab^p \neq 0$, absurd. Therefore we must have $q=r=m$, and then $C_{pm} (P(Q(x)) + P(R(x))) = a(b^p +c^p)\neq 0$, whence $b^p + c^p = 0$, leading to $p$ being odd, and $c=-b$. Consider now $a(Q(x)^p + R(x)^p) = a(Q(x)+R(x)) S(x)$, where we have denoted $ S(x) = Q(x)^{p-1} - Q(x)^{p-2}R(x) + \cdots - Q(x)R(x)^{p-2} + R(x)^{p-1}$. We have $C_{(p-1)m}(S(x)) = b^{p-1} - b^{p-2}(-b) + \cdots - b(-b)^{p-2} + (-b)^{p-1} = pb^{p-1} \neq 0$, so $\deg S(x) \geq (p-1)m$ (in fact it is equal to it). On the other hand, for $T(x) = P(Q(x)) + P(R(x)) - a(Q(x)^p + R(x)^p)$ we have $\deg T(x) \leq (p-1)m$. It follows that assuming $Q(x)+R(x)$ not being constant, thus $\deg(Q(x)+R(x)) \geq 1$, we would have $\deg(a(Q(x)^p + R(x)^p)) = \deg(a(Q(x)+R(x)) S(x)) \geq 1 + (p-1)m$, and so $\deg(P(Q(x)) + P(R(x))) = \deg(a(Q(x)^p + R(x)^p) + T(x)) \geq 1+(p-1)m >0$, absurd. It is also interesting to ask if for any $Q,R$ such that $Q(x) + R(x) = C$ is constant, we can find a non-constant $P$, of any odd degree $p$, such that $P(Q(x)) + P(R(x)) = k$ is constant. Just take $P(x) = (2x-C)^p + k/2$.

IZhO 2012/3 wrote: Let $P, Q,R$ be three polynomials with real coefficients such that \[P(Q(x)) + P(R(x))=\text{constant}\]for all $x$. Prove that $P(x)=\text{constant}$ or $Q(x)+R(x)=\text{constant}$ for all $x$. This is a problem about standard degree and coefficient checking. Assume $P$ and $Q + R$ are both non-constant. It is immediate to check that $\text{deg} \ P \ge 2$. Suppose \[ P(Q(x)) + P(R(x)) = \alpha \]for some constant $\alpha \in \mathbb{R}$. Claim 01. $\text{deg} \ Q = \text{deg} \ R$. Proof. Assume $\text{deg} \ Q \not= \text{deg} \ R$, WLOG $\text{deg} \ Q > \text{deg} \ R$. Therefore \[ 0 = \text{deg} (P \circ Q + P \circ R ) = \text{max}(\text{deg} \ P \circ Q, \text{deg} \ P \circ R) = \text{deg} \ P \circ Q = \text{deg} \ P \cdot \text{deg} \ Q \]This forces $\text{deg} \ P = 0$, which is a contradiction. Claim 02. $\text{deg} \ P$ is odd. Proof. Assume otherwise, that $\text{deg} \ P$ is even. WLOG $P$ has a positive leading coefficient. Therefore $m = \min P$ exists. Since $\text{deg} \ Q = \text{deg} \ R \ge 1$, take really large $x$ such that $P(Q(x)) > \alpha - m$. Therefore, \[ \alpha = P(Q(x)) + P(R(x)) > (\alpha - m) + m = \alpha \]which is a contradiction. Now, write $P(x) = \sum_{i = 0}^n a_i x^i$. Therefore \[ \alpha = \sum_{i = 0}^n a_i (Q(x)^i + R(x)^i) = a_n (Q(x)^n + R(x)^n) + \sum_{i = 0}^{n - 1} a_i (Q(x)^i + R(x)^i) \] Claim 03. $\text{deg} ( Q^{n} + R^{n} ) \ge (n - 1) \cdot \text{deg} \ Q + 1$. Proof. By the previous claim and assumption, $Q + R$ is not a constant, and $n$ is an odd number, and therefore we can write \[ Q^n + R^n = (Q + R)(Q^{n - 1} - Q^{n - 2} R + \dots + R^{n - 1} ) \]We claim that $\text{deg} \ (Q^{n - 1} - Q^{n - 2} R + \dots + R^{n - 1} ) \ge (n - 1) \cdot \text{deg} \ Q$. To prove this, let $[x^{\text{deg} \ Q}] Q = q$ and $[x^{\text{deg} \ R}] R = r$. Therefore, \[ [x^{n + \text{deg} \ Q}] (P \circ Q + P \circ R) = 0 \implies a_n(q^n + r^n) = 0 \]which is enough to forces $q = -r$. Therefore \[ [x^{(n - 1) \text{deg} \ Q}] (Q^{n - 1} - Q^{n - 2} R + \dots + R^{n - 1}) = q^{n - 1} - q^{n - 2} (-q) + \dots + (-q)^{n - 1} = nq^{n - 1} \not= 0 \]which therefore concludes what we wanted, since $\text{deg} (Q + R) \ge 1$. To finish the problem, notice that \begin{align*} 0 &= \text{deg} ( P \circ Q + P \circ R) \\ &= \text{deg} \left( a_n (Q(x)^n + R(x)^n) + \sum_{i = 0}^{n - 1} a_i (Q(x)^i + R(x)^i) \right) \\ &= \text{deg} (a_n (Q(x)^n + R(x)^n)) \\ &\ge (n - 1) \cdot \text{deg} \ Q + 1. \end{align*}since $\text{deg} \left( \sum_{i = 0}^{n - 1} a_i (Q(x)^i + R(x)^i) \right) \le (n - 1) \cdot \text{deg} \ Q$, which is a contradiction. Therefore, our initial assumption is false, and hence, either $P$ or $Q + R$ must be a constant.

Amir Hossein wrote: Let $P, Q,R$ be three polynomials with real coefficients such that \[P(Q(x)) + P(R(x))=\text{constant}\]for all $x$. Prove that $P(x)=\text{constant}$ or $Q(x)+R(x)=\text{constant}$ for all $x$. d is degree of the polynomial, than d[P(Q(x))]=d[P(R(x))] d[P(x)]×d[Q(x)]=d[P(x)]×d[R(x)] d[P(x)]×(d[Q(x)]-d[R(x)])=0 We have d[P(x)]=0 or d[Q(x)]=d[R(x)]=n,(n-natural number) by d[P(x)]=0 we get P(x)=constant. So now we get maximum degree of x:

Suppose the contrary and write $P(Q(x)) + P(R(x)) = c$. Firstly, the polynomials $P(Q(x))$ and $P(R(x))$ must have equal degrees (otherwise their sum would have degree equal to the larger of the two and in particular this degree would be non-zero) and since $\deg P > 0$ (as we supposed that $P$ is non-constant), we obtain $t:= \deg Q = \deg R$. Denote $n = \deg P$ and let $a_n$, $q$, $r$ be the leading coefficients of $P$, $Q$ and $R$, respectively. With $P(x) = \sum_{i=0}^n a_ix^i$ where $a_n \neq 0$ we have \[ c = a_n(Q(x)^n + R(x)^n) + \sum_{i=0}^{n-1}a_i(Q(x)^i + R(x)^i) \]and in particular, the coefficient at the potentially largest degree $nt$ on the right-hand side must be zero. This coefficient is $a_n(q^n + r^n) = 0$, whence $n$ is odd and $q=-r$. The degree of $\sum_{i=0}^{n-1}a_i(Q(x)^i + R(x)^i)$ is bounded from above by the degree of $Q(x)^{n-1} + R(x)^{n-1}$, hence by $t(n-1)$. It remains to argue that $Q(x)^n + R(x)^n$ has degree larger than this upper bound. To do so, relying on $n$ being odd, write \[ Q^n + R^n = (Q+R)(Q^{n-1} - Q^{n-2}R + \cdots - QR^{n-2} + R^{n-1}). \]The degree of the second factor is $(n-1)t$, since the coefficient at $x^{(n-1)t}$ is $q^{n-1} - q^{n-2}r + \cdots - qr^{n-2} + r^{n-1} = nq^{n-1} \neq 0$ (recall $q=-r\neq 0$ from above). The degree of the first factor is at least $1$ (this is the only time we use the assumption that $Q+R$ is non-constant), so the degree of the product overall is larger than $(n-1)t$, which completes the proof.

Assume FTSOC $P$ and $Q+R$ are nonconstant. Shift $P$ so that $P(Q(x))+P(R(x))=0$. This implies $P(Q(x))^2=P(R(x))^2$. Lemma: Let $f$ be a nonconstant even-degree polynomial with positive leading coefficient. For sufficiently large $c$, if $a \ne b$ and $f(a)=f(b)=c$, then $a+b$ is bounded. Proof: Notice that for sufficiently large $c$, the equation $f(x)=c$ has two solutions, one negative and one positive. Assume WLOG $a$ is the negative solution and $b$ is the positive. Scale so that $f$ is monic, then write $f(x)=x^{2n}+a_{2n-1}x^{2n-1}+\cdots$. Choose an integer $k$ with $k-1<\tfrac{a_{2n-1}}{2n}<k+1$. Then, $f(x)$ is bounded between $(x+k-1)^{2n}$ and $(x+k+1)^{2n}$ for large enough. Thus, we have \begin{align*} -\sqrt[2n]{c}-k-1 &\le a \le -\sqrt[2n]{c}-k+1 \\ \sqrt[2n]{c}-k-1 &\le b \le \sqrt[2n]{c}-k+1, \end{align*}so $a+b \in [-2k-2,-2k+2]$, as desired. $\square$ Notice if $Q(x)=R(x)$ holds for some arbitrarily large $x$, then $Q=R$, so $P(Q(x))=0$ for all $x$. If $Q$ is constant, then $Q+R$ is constant. Otherwise, $P$ has infinitely many roots, so it is constant. Thus, we must have $Q(x) \ne R(x)$ for all sufficiently large $x$. The lemma implies $P$ is constant or $Q(x)+R(x)$ is bounded for large enough $x$. However, this implies $Q+R$ is constant, a contradiction. Therefore, $P$ or $Q+R$ is constant, as desired. $\blacksquare$