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p'(x) = nq(x) if p(x)^2 = (x^2 - 1)q(x)^2 + 1

Source: 1978 Swedish Mathematical Competition p6

March 26, 2021
analysisalgebrapolynomial

Problem Statement

p(x)p(x) is a polynomial of degree nn with leading coefficient cc, and q(x)q(x) is a polynomial of degree mm with leading coefficient cc, such that p(x)2=(x2−1)q(x)2+1 p(x)^2 = \left(x^2 - 1\right)q(x)^2 + 1 Show that p′(x)=nq(x)p'(x) = nq(x).
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