My solution: note that $(P(x)+Q(x))(P(x)^2+Q(x)^2-P(x)Q(x))=(x^4+1)(x^8-x^4+1)$ it is easy to observe that $(x^4+1,x^8-x^4+1)=1,(P(x)+Q(x),P(x)^2+Q(x)^2-P(x)Q(x))=1$ and $x^4+1,x^8-x^4+1$ are irreducible. so there are two cases: $1)P(x)+Q(x)=x^4+1,P(x)^2+Q(x)^2-P(x)Q(x)=x^8-x^4+1\longrightarrow (x^4+1)^2-3P(x)Q(x)=x^8-x^4+1\longrightarrow P(x)Q(x)=x^4$ so $P(x)+Q(x)=x^4+1,P(x)Q(x)=x^4\longrightarrow \boxed{1: P(x)=x^4,Q(x)=1},\boxed{1: P(x)=1,Q(x)=x^4}$ $2)P(x)+Q(x)=x^8-x^4+1,P(x)^2+Q(x)^2-P(x)Q(x)=x^4+1$ note that for large value of $x$ we have $x^8-x^4+1>x^4+1\longrightarrow P(x)+Q(x)>P(x)^2+Q(x)^2-P(x)Q(x)\longrightarrow 2>(P(x)-Q(x))^2+(P(x)-1)^2+(Q(x)-1)^2$ but it's clearly false for large value of $x$ contradiction. DONE(please someone correct me if i am wrong)

Sorry, but i have some questions... <andria> wrote: $x^4+1,x^8-x^4+1$ are irreducible. How could you know that? (well... I'm not good at polynomials)

@above Cause they are cyclotomic polynomials.

@above But every polynomial of degree $3$ or more is reducible. It either has a real root $r$ or two complex roots that are conjugates $c$ and $\bar{c}$. So it is either divisible by $x-r$ or $x^2-x(c+\bar{c})+c\cdot\bar{c} \in \mathbb{R}[x]$ UPD. Yeah, didn't notice $\mathbb{Q}[x]$

ComiCabE wrote: @above But every polynomial of degree $3$ or more is reducible. It either has a real root $r$ or two complex roots that are conjugates $c$ and $\bar{c}$. So it is either divisible by $x-r$ or $x^2-x(c+\bar{c})+c\cdot\bar{c} \in \mathbb{R}[x]$ Except that you didn't see the $\mathbb{Q}[x]$

Solved with nukelauncher. Observe the factorizations \begin{align*} x^{12}+1&=\left(x^4+1\right)\left(x^8-x^4+1\right)\\ P^3+Q^3&=(P+Q)\left(P^2+Q^2-PQ\right). \end{align*}It can be verified that \(x^{12}+1\) is fully factorized above; hence \[\left\{P+Q,P^2+Q^2-PQ\right\}=\left\{k\left(x^4+1\right),k^{-1}\left(x^8-x^4+1\right)\right\}\]for some constant \(k\). However \(\deg(P^2+Q^2-PQ)=2\max\{\deg P,\deg Q\}>\deg(P+Q)\). (Note that if \(\deg P=\deg Q\), then the leading coefficient of \(P^2+Q^2-PQ\) is still positive.) Hence \begin{align*} P+Q&=k\left(x^4+1\right)\\ P^2+Q^2-PQ&=k^{-1}\left(x^8-x^4+1\right). \end{align*} Claim: \(\deg P\ne\deg Q\). Proof. Assume not, and let \(p\), \(q\) be the leading coefficients of \(P\), \(Q\). The leading coefficient \(p^2+q^2-pq>0\) does not vanish in \(P^2+Q^2-PQ\), so \(\deg P=\deg Q=4\). Then by \(P^3+Q^3=x^{12}+1\), we have \[p^3+q^3=1,\]which is impossible for rational \(p\), \(q\) by Fermat's last theorem. \(\blacksquare\) Then let \(\deg P=4\) and \(\deg Q<4\). so the leading coefficient of \(P+Q\) is \(p=k\), and the leading coefficient of \(P^2+Q^2-PQ\) is \(p^2=k^{-1}\); it follows that \(p=k=1\). Hence \begin{align*} P+Q&=x^4+1\\ P^2+Q^2-PQ&=x^8-x^4+1. \end{align*}Then observe that \[x^8-x^4+1=(P+Q)^2-3PQ=\left(x^8+2x^4+1\right)-3PQ,\]hence \(PQ=x^4\). The only solution is \(P(x)\equiv x^4\), \(Q\equiv1\) (and its permutation), which works.

At the $$ TheUltimate123 wrote: Solved with nukelauncher. Observe the factorizations \begin{align*} x^{12}+1&=\left(x^4+1\right)\left(x^8-x^4+1\right) P^3+Q^3&=(P+Q)\left(P^2+Q^2-PQ\right). \end{align*} It can be verified that \(x^{12}+1\) is fully factorized above; hence \[\left\{P+Q,P^2+Q^2-PQ\right\}=\left\{k\left(x^4+1\right),k^{-1}\left(x^8-x^4+1\right)\right\}\] for some constant \(k\). However \(\deg(P^2+Q^2-PQ)=2\max\{\deg P,\deg Q\}>\deg(P+Q)\). (Note that if \(\deg P=\deg Q\), then the leading coefficient of \(P^2+Q^2-PQ\) is still positive.) Hence \begin{align*} P+Q&=k\left(x^4+1\right) P^2+Q^2-PQ&=k^{-1}\left(x^8-x^4+1\right). \end{align*} Claim: \(\deg P\ne\deg Q\). Proof. Assume not, and let \(p\), \(q\) be the leading coefficients of \(P\), \(Q\). The leading coefficient \(p^2+q^2-pq>0\) does not vanish in \(P^2+Q^2-PQ\), so \(\deg P=\deg Q=4\). Then by \(P^3+Q^3=x^{12}+1\), we have \[p^3+q^3=1,\] which is impossible for rational \(p\), \(q\) by Fermat's last theorem. \(\blacksquare\) Then let \(\deg P=4\) and \(\deg Q<4\). so the leading coefficient of \(P+Q\) is \(p=k\), and the leading coefficient of \(P^2+Q^2-PQ\) is \(p^2=k^{-1}\); it follows that \(p=k=1\). Hence \begin{align*} P+Q&=x^4+1 P^2+Q^2-PQ&=x^8-x^4+1. \end{align*} Then observe that \[x^8-x^4+1=(P+Q)^2-3PQ=\left(x^8+2x^4+1\right)-3PQ,\] hence \(PQ=x^4\). The only solution is \(P(x)\equiv x^4\), \(Q\equiv1\) (and its permutation), which works. so like can you use fermats last theorem at the real test??

David_Kim_0202 wrote: At the $$ TheUltimate123 wrote: Solved with nukelauncher. Observe the factorizations \begin{align*} x^{12}+1&=\left(x^4+1\right)\left(x^8-x^4+1\right) P^3+Q^3&=(P+Q)\left(P^2+Q^2-PQ\right). \end{align*} It can be verified that \(x^{12}+1\) is fully factorized above; hence \[\left\{P+Q,P^2+Q^2-PQ\right\}=\left\{k\left(x^4+1\right),k^{-1}\left(x^8-x^4+1\right)\right\}\] for some constant \(k\). However \(\deg(P^2+Q^2-PQ)=2\max\{\deg P,\deg Q\}>\deg(P+Q)\). (Note that if \(\deg P=\deg Q\), then the leading coefficient of \(P^2+Q^2-PQ\) is still positive.) Hence \begin{align*} P+Q&=k\left(x^4+1\right) P^2+Q^2-PQ&=k^{-1}\left(x^8-x^4+1\right). \end{align*} Claim: \(\deg P\ne\deg Q\). Proof. Assume not, and let \(p\), \(q\) be the leading coefficients of \(P\), \(Q\). The leading coefficient \(p^2+q^2-pq>0\) does not vanish in \(P^2+Q^2-PQ\), so \(\deg P=\deg Q=4\). Then by \(P^3+Q^3=x^{12}+1\), we have \[p^3+q^3=1,\] which is impossible for rational \(p\), \(q\) by Fermat's last theorem. \(\blacksquare\) Then let \(\deg P=4\) and \(\deg Q<4\). so the leading coefficient of \(P+Q\) is \(p=k\), and the leading coefficient of \(P^2+Q^2-PQ\) is \(p^2=k^{-1}\); it follows that \(p=k=1\). Hence \begin{align*} P+Q&=x^4+1 P^2+Q^2-PQ&=x^8-x^4+1. \end{align*} Then observe that \[x^8-x^4+1=(P+Q)^2-3PQ=\left(x^8+2x^4+1\right)-3PQ,\] hence \(PQ=x^4\). The only solution is \(P(x)\equiv x^4\), \(Q\equiv1\) (and its permutation), which works. so like can you use fermats last theorem at the real test?? In the words of Ali Adali, IMO Bronze medalist (and my NT course teacher), "Fermat's Last Theorem is not a good thing to cite on real olympiad. The graders expect that if you cite a theorem, you know how to prove it, and citing Fermat's Last theorem wont give you points..." In my opinion - it should be perfectly fine but apparently it's not

Amkan2022 wrote: David_Kim_0202 wrote: At the $$ TheUltimate123 wrote: Solved with nukelauncher. Observe the factorizations \begin{align*} x^{12}+1&=\left(x^4+1\right)\left(x^8-x^4+1\right) P^3+Q^3&=(P+Q)\left(P^2+Q^2-PQ\right). \end{align*} It can be verified that \(x^{12}+1\) is fully factorized above; hence \[\left\{P+Q,P^2+Q^2-PQ\right\}=\left\{k\left(x^4+1\right),k^{-1}\left(x^8-x^4+1\right)\right\}\] for some constant \(k\). However \(\deg(P^2+Q^2-PQ)=2\max\{\deg P,\deg Q\}>\deg(P+Q)\). (Note that if \(\deg P=\deg Q\), then the leading coefficient of \(P^2+Q^2-PQ\) is still positive.) Hence \begin{align*} P+Q&=k\left(x^4+1\right) P^2+Q^2-PQ&=k^{-1}\left(x^8-x^4+1\right). \end{align*} Claim: \(\deg P\ne\deg Q\). Proof. Assume not, and let \(p\), \(q\) be the leading coefficients of \(P\), \(Q\). The leading coefficient \(p^2+q^2-pq>0\) does not vanish in \(P^2+Q^2-PQ\), so \(\deg P=\deg Q=4\). Then by \(P^3+Q^3=x^{12}+1\), we have \[p^3+q^3=1,\] which is impossible for rational \(p\), \(q\) by Fermat's last theorem. \(\blacksquare\) Then let \(\deg P=4\) and \(\deg Q<4\). so the leading coefficient of \(P+Q\) is \(p=k\), and the leading coefficient of \(P^2+Q^2-PQ\) is \(p^2=k^{-1}\); it follows that \(p=k=1\). Hence \begin{align*} P+Q&=x^4+1 P^2+Q^2-PQ&=x^8-x^4+1. \end{align*} Then observe that \[x^8-x^4+1=(P+Q)^2-3PQ=\left(x^8+2x^4+1\right)-3PQ,\] hence \(PQ=x^4\). The only solution is \(P(x)\equiv x^4\), \(Q\equiv1\) (and its permutation), which works. so like can you use fermats last theorem at the real test?? In the words of Ali Adali, IMO Bronze medalist (and my NT course teacher), "Fermat's Last Theorem is not a good thing to cite on real olympiad. The graders expect that if you cite a theorem, you know how to prove it, and citing Fermat's Last theorem wont give you points..." In my opinion - it should be perfectly fine but apparently it's not Thanks you!!