P(x)P(x) is polynomial with integer coefficients and ww is complex (primitive) root of unity of a degree nn. Show that P(1)∗P(w)∗P(w2)∗…∗P(wn−1)P(1)*P(w)*P(w^2)*…*P(w^{n-1}) is integer number for every natural number nn.

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1 Answer

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Let wk=wkw_k=w^k. Then Q(w0,w1,…,wn−1)=P(w0)⋅P(w1)⋅P(w2)⋯P(wn−1)Q(w_0, w_1, \dots,w_{n-1})=P(w_0) \cdot P(w_1) \cdot P(w_2) \cdots P(w_{n-1}) is a polynomial symmetric in w0,w1,…,wn−1w_0, w_1, \dots,w_{n-1}. Therefore QQ can be expressed as a polynomial in the elementary symmetric polynomials in wkw_k with integer coefficients. But, since ww is a primitive nthn^{th} root of unity, wk=wk|k=0…n−1w_k=w^k \;|\; k=0 \dots n-1 are the nn roots of xn−1=0x^n – 1=0 which is a monic polynomial with integer coefficients, so by Vieta’s formulas all the elementary symmetric polynomials in wkw_k are integers (in fact, all of them 00 except the full product which is 11), thus QQ is an integer as well.