Home Competitions USAMO 1992 Problem 5
A complex polynomial has degree 1992 and distinct zeros. Show that we can find complex numbers z n \mathrm{z}_{\mathrm{n}} z n p 1 ( z ) = z − z 1 \mathrm{p}_{1}(\mathrm{z})=\mathrm{z}-\mathrm{z}_{1} p 1 ( z ) = z − z 1 p n ( z ) = p n − 1 ( z ) 2 − z n \mathrm{p}_{\mathrm{n}}(\mathrm{z})=\mathrm{p}_{\mathrm{n}-1}(\mathrm{z})^{2}-\mathrm{z}_{\mathrm{n}} p n ( z ) = p n − 1 ( z ) 2 − z n p 1992 ( z ) \mathrm{p}_{1992}(\mathrm{z}) p 1992 ( z )
Dataset text NuminaMath olympiads . Quoted with source.
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