olymmath

KöMaL 2010 A511

    Show that for arbitrary positive integers nn, kk there exists a polynomial pp(xx), with degree at most 100nk100\sqrt{nk} , such that p(0)>(∣p(1)∣+…+∣p(n)∣)+(∣p(−1)∣+…+∣p(−k)∣)p(0) > \big(\big|p(1)\big|+\ldots+\big|p(n)\big|\big) + \big(\big|p(-1)\big|+\ldots+\big|p(-k)\big|\big).

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