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Question Number 1844 by 123456 last updated on 12/Oct/15

lets two polynimies p_n ,q_n givwn by p_1 =q_1 =x p_(n+1) =p_n +q_n q_(n+1) =p_n q_n then (1,1)→(1,2)→(2,3) lets W(u,v)= determinant ((u,v),((u′),(v′))) is true that W(p_n ,q_n )≠0,∀n>1

$$\mathrm{lets}\:\mathrm{two}\:\mathrm{polynimies}\:{p}_{{n}} ,{q}_{{n}} \:\mathrm{givwn}\:\mathrm{by} \\ $$$${p}_{\mathrm{1}} ={q}_{\mathrm{1}} ={x} \\ $$$${p}_{{n}+\mathrm{1}} ={p}_{{n}} +{q}_{{n}} \\ $$$${q}_{{n}+\mathrm{1}} ={p}_{{n}} {q}_{{n}} \\ $$$$\mathrm{then}\:\left(\mathrm{1},\mathrm{1}\right)\rightarrow\left(\mathrm{1},\mathrm{2}\right)\rightarrow\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\mathrm{lets}\:\mathrm{W}\left({u},{v}\right)=\begin{vmatrix}{{u}}&{{v}}\\{{u}'}&{{v}'}\end{vmatrix} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{that} \\ $$$$\mathrm{W}\left({p}_{{n}} ,{q}_{{n}} \right)\neq\mathrm{0},\forall{n}>\mathrm{1} \\ $$

Commented by Rasheed Soomro last updated on 12/Oct/15