Let α,β and γ be the distinct integer zeroes of the polynomial P. If possible let P(m)=1 where m∈Z. Since P∈Z(x)
we have α−m|P(α)−P(m)⟹α−m|(−1). Similarly β−m|(−1) and γ−m|(−1). Since α,β and γ are distinct α−m,β−m and γ−m are distinct. This shows that α−m,β−m and γ−m are distinct factors of 1, which is impossible! So the assumption that P(m)=1 is not tenable for any integer m.
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Saturday, June 13, 2015
Indian Statistical Institute B.Math & B.Stat : Polynomials
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