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Algebra Difficulty 5.7 AIME, harder Prove it

13. (Euclidean algorithm in Q[x]Q[x]) Let f0,f1Q[x]f_{0}, f_{1} \in Q[x], f10,f1f0f_{1} \neq 0, f_{1} \nmid f_{0}. Prove:
(i) It is always possible to repeatedly apply the division algorithm from the previous problem to obtain the following k+1k+1 equations:
f0=q0f1+f2,degf2<degf1,f20,f1=q1f2+f3,degf3<degf2,f30,fk1=qk1fk+fk+1,fk=qkfk+1;degfk+1<degfk,fk+10,\begin{array}{ll} f_{0}=q_{0} f_{1}+f_{2}, & \operatorname{deg} f_{2}<\operatorname{deg} f_{1}, f_{2} \neq 0, \\ f_{1}=q_{1} f_{2}+f_{3}, & \operatorname{deg} f_{3}<\operatorname{deg} f_{2}, f_{3} \neq 0, \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ f_{k-1}=q_{k-1} f_{k}+f_{k+1}, & \cdots \cdots \cdots \cdots \cdots \cdots \\ f_{k}=q_{k} f_{k+1} ; & \operatorname{deg} f_{k+1}<\operatorname{deg} f_{k}, f_{k+1} \neq 0, \end{array}
(ii) fk+1fj(0jk)f_{k+1} \mid f_{j}(0 \leqslant j \leqslant k);
(iii) For each j(0j<k)j(0 \leqslant j<k), there exist hj,hj+1Q[x]h_{j}, h_{j+1} \in Q[x], such that
fk+1=hjfj+hj+1fj+1.f_{k+1}=h_{j} f_{j}+h_{j+1} f_{j+1} .

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.