This one was a really good free problem. WLOG $deg(P)>deg(Q),deg(R)$, then from $$P+Q+R=0$$we'll get that $P=-(Q+R)$ which is a contradiction since they're non-zero. Then WLOG $deg(Q)=deg(P)=n$, $deg(R)=m$. But it's easy to see that from $$P(Q)+Q(R)+R(P)=0$$we get: $$P(Q)=-(Q(R)+R(P)) \Rightarrow n^2=mn$$contradiction for $m \neq n$, so $n=m$. Which means all $3$ polynomials have equal degrees. Since $P+Q+R=0$ we have that the sum of their leading coefficients have to be $0$, so we can take the leading coefficients of $P$, $Q$, and $R$ as $p_n$, $q_n$, and $-p_n-q_n$ respectively. For the second set of polynomials as well, so we get the sum of the leading coefficients of $P(Q)+Q(R)+R(P)=0$ is also $0$. So we get the following equality: $$p_n \cdot q_n^n+q_n(-p_n-q_n)^n+(-p_n-q_n)p_n^n=0$$Now if $ 2 \nmid n$ we can take the negative signs out and have the following equality: $$p_n \cdot q_n^n = q_n(p_n+q_n)^n + (p_n+q_n)p_n^n$$Now if $p_n$ and $q_n$ were to have the same signs it's obvious that the RHS would be bigger, so they have different signs, which is now just a case bash: 1)$p_n<0$ $q_n>0 \Rightarrow p_n=-k$ 1.1)$q_n>k$ Substituting $p_n=-k$, we'll get $$q_n(q_n-k)^n+kq_n^n=(k_n-k)k^n \Rightarrow$$$$\Rightarrow q_n(q_n-k)^n+kq_n^n+k^{n+1}=k^nq_n$$But $kq_n^n>k^nq_n$, contradiction. 1.2) $q_n<k$ We change some signs and get: $$-qn(k-q_n)^n+kq_n^n=-(k-q_n)k^n \Rightarrow$$$$\Rightarrow kq_n^n+(k-q_n)k^n=q_n(k-q_n)^n$$But $k^n(k-q_n)>q_n(k-q_n)^n \iff k^n > q_n(k-q_n)^{n-1}$ Which is true by $k>q_n$. 2)$q_n<0$, $p_n >0 \Rightarrow q_n=-k$ 2.1)$p_n>k$ Substituting $q_n=-k$, we get: $$(p_n-k)p_n^n+k^np_n=k(p_n-k)^n$$But $$(p_n-k)p_n^n>k(p_n-k)^n \iff$$$$\iff p_n^n > k(p_n-k)^{n-1}$$Which is true by $p_n>k$. 2.2) $k>p_n$ $$-(k-p_n)p_n^n+k^np_n=-k(k-p_n)^n \Rightarrow$$$$\Rightarrow k(k-p_n)^n+k^np_n=(k-p_n)p_n^n \Rightarrow$$$$\Rightarrow k(k-p_n)^n + p_n^{n+1}+k^np_n=kp_n^n$$But $k^np_n>kp_n^n$. Contradiction. Exhausting all cases we see that $n$ can't be odd, which means it's even. Q.E.D Sultanbek

We are claiming that $degP=degQ=degR$. We will prove this using following lemma. Lemma : if $P_1(x)+P_2(x)+P_3(x)=0$ then two polynomials have same degree and It is greater (or equal) than 3rd one. proof : Assume contradiction. Lets say $P_1(x)$ has a maximum degree and assume that its leading coefficient is positive. Then by putting large $x$ we have $P_1(x)+P_2(x)+P_3(x)>0$ contradiction So using lemma we have $degP=degQ=n$ and $degR=k\leq n$. Also using lemma we can say that $degP(Q(x))=deg(Q(R(x))$ or $degP(Q(x))=deg(R(P(x))$.Either way $n=k$. Let say leading coefficients of polynomials are $p,q,r$. Assume contradiction. Let say $n$ is odd. It's easy to see that $p+q+r=0$ and $pq^n+qr^n+rp^n=0$. WLOG $p,q>0$ and $r<0$. Let $s=-r$ then we have $p+q=s$ and $pq^n=qs^n+sp^n=qs^n+qp^n+p^{n+1}$. Since $pq^n>qp^n$ we have $q>p$. On the other hand $s>q$ therefore $qs^n>pq^n$ contradiction.

The second part after proving the equal degrees is the same as in #2 and #3, so I won't bother typing it again. I will just sketch my proof for the equal degrees. The idea is to prove that $deg(P) \leq deg(Q)$, similar inequalities will follow cyclically and we will obtain what we want. To prove this, suppose FTSoC $deg(P)>deg(Q)$, and note that the second equation rewrites as $P(Q(x))-Q(P(x))=P(P(x))-Q(-P(x)-Q(x))$ and just note that the LHS has deg at most $deg(P).deg(Q)$, and the RHS has degree $deg(P)^2$, contradiction to the assumption.

Unexpectedly many technical details (around simple ideas though) for P1, but nevertheless it is a great problem! We firstly show that $\deg P = \deg Q = \deg R$. Note that $P(Q(x)) + Q(-P(x)-Q(x)) = P(P(x)) + Q(P(x))$ and here the degrees of the four compositions are $\deg P \cdot \deg Q$, $\deg Q \cdot \max(\deg P, \deg Q)$, $(\deg P)^2$ and $\deg P \cdot \deg Q$. So if $\deg P < \deg Q$, then the degree on the left is $(\deg Q)^2$ while on the right it is $\deg P \cdot \deg Q$, whence $\deg P = \deg Q$, contradiction. If $\deg P > \deg Q$, then the degree on the left is $\deg P \cdot \deg Q$ while on the right it is $(\deg P)^2$, whence $\deg P = \deg Q$, contradiction. Therefore $\deg P = \deg Q$ and analogously $\deg Q = \deg R$, so $\deg P = \deg Q = \deg R = k$. Our aim is to show that $k$ cannot be odd, so suppose it is indeed odd. Let $a$, $b$ and $c$ be the leading coefficients of $P$, $Q$ and $R$ -- from $P+Q+R = 0$ we get $a+b+c=0$ and from $P(Q(x)) + Q(R(x)) + R(P(x)) = 0$ we get $ab^k + bc^k + ca^k = 0$. Hence $c=-a-b$ and $ab^k = b(a+b)^k + (a+b)a^k$, now with $a=tb$ and dividing by $b^{k+1}$ we get $t = t^k(t+1) + (t+1)^k$, i.e. $(t+1)^k + t^{k}(t+1) - t = 0$. It remains to show that the latter has no real roots. If $t\geq 0$, then $(t+1)^k > t$ and $t^{k+1} > 0$, $t^k > 0$. For $t<0$ let $z=-t > 0$ and consider $(z-1)^k = z^k(z-1) + z$ (we have used that $k$ is odd). If $z>1$, then $z^k > (z-1)^k > (z-1)^{k-1}$ and so the right-hand side is larger. Clearly $z=1$ is impossible. Finally, if $z\in (0,1)$, then in the left-hand side is negative, while the right-hand side is positive since $z > z^k$ implies $z^k(z-1) + z > z^k(z-1) > z^k = z^{k+1} > 0$.

Obviously we must have that the degrees of two polynomials must be the same, assume that $\text{deg}(P)=\text{deg}(R)$, and by the right hand side of the relation we have that $\text{deg}(Q)=\text{deg}(P) = \text{deg}(R) = n$. Thus we have that: $$P=\sum_{i=0}^{n}a_ix^i$$$$Q=\sum_{i=0}^{n}b_ix^i$$$$R=\sum_{i=0}^{n}c_ix^i$$from the first relation we then must have that $R=-\sum_{i=0}^n(a_i+b_i)x^i$ Now assume that $n \equiv 1 \pmod{2}$. From the second relation we must have that: $$\sum a_iQ(x)^i + b_iR(x)^i = \sum (a_i+b_i)P(x)^i$$plugging in $R$ into this we have that the coefficient placed on $x^{n^2}$, is: $$ab^n-b^{n+1}-ba^n = a^{n+1}+ba^n$$rearranging we get that: $$a^{n+1}+b^{n+1}=ab(b^{n-1}-2a^{n-1}) > 0$$but notice that this implies that: $$ab^n+a^nb >a^{n+1}+b^{n+1}$$which isn't true obviously (the upper inequality must be analysed into four cases $b < 0, b^{n-1}-2a^{n-1} < 0$, $a <0, b^{n-1}-2a^{n-1} < 0$, $a > 0, b > 0$ and $a < 0, b < 0$). Thus we must have that $n \equiv 0 \pmod{2}$

mathematics2004 wrote: Non-zero polynomials $P(x)$, $Q(x)$, and $R(x)$ with real coefficients satisfy the identities $$ P(x) + Q(x) + R(x) = P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$Prove that the degrees of the three polynomials are all even. $$P(x)= a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0$$$$Q(x)= b_m \cdot x^m + ... + b_2 \cdot x^2 + b_1 \cdot x + b_0$$$$a_n,b_m \neq 0$$because $P(x)$ and $Q(x)$ are polynomials different from $0$. Suppose $m> n$. $$R(x) = -P(x)-Q(x) \implies deg R =m$$$$P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$$$deg(P(Q(x))=mn,deg(Q(R(x))= m^2,deg(R(P(x)) = mn$$Note that $m^2> mn$ .Then the coefficient in front of $m^2$ should be $0$. we calculate the coefficient before $x^{m^2}$. $P(Q(x)) + R(P(x))$ does not include $x^{m^2}$. $$Q(R(x))= b_m(R(x))^m + b_{m-1}(R(x))^{m-1} + ... + b_0 $$So the coefficient before $x^{m^2}$ is $b_m \cdot (-b_m)^m$ $\implies b_m \cdot (-b_m)^m = 0$.Contradiction. a similar $n>m$ case is also a contradiction. So $degP=degQ$.$\implies$ similarity $degP = deg Q$. So we have the following. $$\boxed{degP=deg Q =deg R}$$Let's say $n$ is an odd number. $$P(x)= a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0$$$$Q(x)= b_n \cdot x^n + ... + b_2 \cdot x^2 + b_1 \cdot x + b_0$$$$R(x)= c_n \cdot x^n + ... + c_2 \cdot x^2 + c_1 \cdot x + c_0$$ $$a_n=a,b_n=b,c_n=c,a+b+c=0$$The coefficient in front of $x^{n^2}$ is $0$.So we have the following expression. $$\boxed{ab^n + bc^n + ca^n = 0}$$at least 2 of $a,b,c$ have the same sign.Suppose $a$ and $b$. $1)a,b > 0 \implies ab^n = b(a+b)^n+(a+b)a^n > a^{n+1}+b^{n+1} \implies $ Contradiction. $2)a,b<0 $This is the same as above.So $n$ is a evan number.$\blacksquare$

By the first condition, $R(x)=-P(x)-Q(x)$ for all $x$. Using the second condition, $$P(Q(x))+Q(-P(x)-Q(x))=P(P(x))+Q(Q(x)) (*)$$. Let $D(T(x))$ denote the degree of the polynomial $T(x)$. Claim : $D(P(x))=D(Q(x))$ Let the quantities be $m$ and $n$ respectively. if $m>n$, then comparing the degrees in $(*)$, $mn \geq m^2 \implies n \geq m$ if $n>m$, then comparing the degrees as before, $n^2=mn \implies m=n$ Now for the sake of contradiction, let the degree of $P(x)$ be $2k-1$ for some $k \in \mathbb{N}$ Let $P(x)=\sum{a_ix^i}$ and $Q(x)=\sum{b_ix^i}$ where $a_{2k-1} \neq 0, b_{2k-1} \neq 0$ Now we compute the coefficient of $x^{(2k-1)^2}$ in the second condition to get : (Let $2k-1=m$ for ease) $$a_mb_m^m=b_m(a_m+b_m)^m+a_m^m(a_m+b_m)$$Divide both sides by $b_m^{m+1}$ and let $t=\frac{a_m}{b_m}$ $$t=(t+1)^m+t^{m+1}+t^m$$ If $t>0$, this is trivial if $t<0$, replace $t$ by $-T$ where $T>0$ to get $T+T^{2k}=(T-1)^{2k-1}+T^{2k-1}$ if $T>1$, replace $T=1+S$ and use binomial to finish if $1>T>0$, replace $T$ by $\frac{1}{x}$ for $x>1$ to get $x^{2k-1}+1+x(x-1)^{2k-1}=x$ which is obviously not possible. With all these cases done, we know that this equation can never be true, i.e. $m \neq 2k-1$ Now if $a_m+b_m \neq 0$, we would have $a_mb_m^m=0$ which is not possible, i.e. degree of $R$ must be the same as that of $P$ and $Q$, and therefore even $\blacksquare$ Remarks: This is needlessly complicated. One could have WLOG assumed that the sign of $P$ and $Q$ is the same, getting rid of all of that casework.

According to the first condition, it is clear that two polynomials have the same degree, and the third degree has less of them (otherwise, the maximum degree cannot be reduced in the equation). If you look at the second equation, you can understand that the degrees of all polynomials are equal. Let's assume that all polynomials are odd. Let's say $a, b, c$ are the coefficients of these higher powers in the polynomial respectively, which means $a+b+c=0$, and according to the second equation it follows that $ab^n + bc^n +ca^n=0$. Also, some two numbers have the same sign, let $a,b>0$ and $c=-(a+b)<0$, it also follows that $|c|>a, b>0$. So $|bc^n +ca^n|= ab^n$. But $|bc^n +ca^n|> |bc^n|> b*|c|*|c^{n-1}|> a*b*b^{n-1}$ which is a contradiction. For the case of two negative ones, the same.

Claim : All polynomials $P(x)$, $Q(x)$, and $R(x)$ have same degree $n$. Proof : Let assume maximum of $deg(P),deg(Q),deg(R)$ is $n$. Let $a_1,a_2,a_3$ be coefficients of $x^n$ in polynomials (at least one of them is nonzero). we have $a_1 + a_2 + a_3 = 0$ and $a_1a_2^n + a_2a_3^n + a_3a_1^n = 0$. if $2$ of $a_1,a_2,a_3$ are $0$ then the third one will be $0$ as well which gives contradiction. WLOG assume $a_3 = 0$ then $a_1a_2^n = 0$ which implies another on is $0$ as well which gives contradicition so $a_1,a_2,a_3 \neq 0$. Let $n$ be odd. between $a_1,a_2,a_3$ two have same sign for example $a_1,a_2$ so $a_3 = -(a_1 + a_2) < 0$ but then $|a_2a_3^n +a_3a_1^n|= a_1a_2^n$ and $|a_2a_3^n +a_3a_1^n| > |a_2a_3^n| > a_1a_2^n$ so contradiction.

The idea is looking at how polynomials increase/decrease over $x$, which is equivalent to checking leading coefficients.

Claim: degrees of all three polynomials are equal. Proof: Let the smallest degree among three polynomials be $m$, and $W.L.O.G.$ say $deg(R(x))=m$. From the first equation $$deg(P(x))=deg(-Q(x)-R(x))=max(deg(Q(x)),deg(R(x)))=deg(Q(x))\Rightarrow deg(P(x))=deg(Q(x)).$$Now from the second equation $$deg(Q(R))=deg(-P(Q(x))-R(P(x)))=max(deg(P(Q(x))),deg(R(P(x))))=deg(P(Q(x)))\Rightarrow deg(P(x))=deg(R(x))$$because $deg(Q(x))\geq deg(R(x))$. The two results above combined imply the claim. $\square$ Claim: m can't be odd. Proof: Assume $m$ is odd. First equation implies that sum of leading coefficients of three polynomials is $0$, so two of those leading coefficient have the same sign and third has a sign different from other two. Now $W.L.O.G.$ let the leading coefficients of $P(x),Q(x),R(x)$ be $a,b$ and $-(a+b)$, respectively with $a,b>0$. From second equation it follows that $$(a(b)^m)+(b(-a-b)^m)+(-(a+b)(a)^m)=0\Rightarrow ab^m=b(a+b)^m+(a+b)a^m>b(a+b)^m>b(ab^{m-1})=ab^m$$Which is a contradiction. $\square$ From above claims it is evident that all three polynomials have even degree.

If exactly one polynomial has maximal degree, we get a contradiction from $P+Q+R=0$. If exactly two polynomials have maximal degree, we get a contradiction from $P \circ Q+Q \circ R+R\circ P=0$, since $\deg(P \circ Q)=\deg P \deg Q$. Thus all three polynomials have the same degree $n$. Suppose $n$ is odd. Let the leading coefficients of $P,Q,R$ be $p,q,r \in \mathbb{R}$ respectively. Then we should have $p+q+r=0$ and $pq^n+qr^n+rp^n=0$. I claim that these two equations have no solutions over $\mathbb{R}$ where no variables are zero (when $n$ is odd). Since $p+q+r=0$, two variables must be the same sign with the third being opposite. WLOG let $p,r$ have the same sign and $|p|<|r|$. Since both equations are homogeneous, scale so that $p=1 \implies r>1$. Then $q=-1-r$, and by substituting we have $$(-1-r)^n+(-1-r)r^n+r=0 \implies r=(r+1)^n+(r+1)r^n,$$which is absurd since $(r+1)^n+(r+1)r^n>(r+1)r^n>r^n>r$. $\blacksquare$