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Omni-MATH / Let F_(m) be the mth Fibonacci number, defined by F₁=F₂=1 and F_(m)=F_(m−1)+F_(m−2) for all m≥3. Let p(x) be the polynomial of degree 1008 such that p(2n+1)=F_(2n+1) for …

Problem

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problem

Let FmF_m be the mmth Fibonacci number, defined by F1=F2=1F_1 = F_2 = 1 and Fm=Fm−1+Fm−2F_m = F_{m-1} + F_{m-2} for all m≥3m \geq 3. Let p(x)p(x) be the polynomial of degree 10081008 such that p(2n+1)=F2n+1p(2n+1) = F_{2n+1} for n=0,1,2,…,1008n=0,1,2,\dots,1008. Find integers jj and kk such that p(2019)=Fj−Fkp(2019) = F_j - F_k.
Plain-text mathematical notation (without MathML)
Let F_(m) be the mth Fibonacci number, defined by F₁=F₂=1 and F_(m)=F_(m−1)+F_(m−2) for all m≥3. Let p(x) be the polynomial of degree 1008 such that p(2n+1)=F_(2n+1) for n=0,1,2,…,1008. Find integers j and k such that p(2019)=F_(j)−F_(k).
Original LaTeX notation
Let $F_m$ be the $m$th Fibonacci number, defined by $F_1 = F_2 = 1$ and $F_m = F_{m-1} + F_{m-2}$ for all $m \geq 3$. Let $p(x)$ be the polynomial of degree $1008$ such that $p(2n+1) = F_{2n+1}$ for $n=0,1,2,\dots,1008$. Find integers $j$ and $k$ such that $p(2019) = F_j - F_k$.

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Source and history

task_id
b7a1ede8-9361-56fd-bc62-766c692fc87c
task_key
test--b7a1ede8-9361-56fd-bc62-766c692fc87c
task_revision_id
1
config / split
/ test

Official source

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