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Math Help - A Problem of Polynomial

  1. #1
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    Thumbs up A Problem of Polynomial

    Give a positive integer n>1. Let f_{i}(x) = a_{i}x+b_{i},i = 1,2,3,nonzero,real polynomial. we have {f_{1}(x)}^n + {f_{2}(x)}^n = {f_{3}(x)}^n . Show that there exist a real polynomial such that f_{i}(x) = c_{i}f(x),where c_{1},c_{2},c_{3} is real numbers.

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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Give a positive integer n>1. Let f_{i}(x) = a_{i}x+b_{i},i = 1,2,3,nonzero,real polynomial. we have {f_{1}(x)}^n + {f_{2}(x)}^n = {f_{3}(x)}^n . Show that there exist a real polynomial such that f_{i}(x) = c_{i}f(x),where c_{1},c_{2},c_{3} is real numbers.
    Here's an outline of one way to do it. (I haven't checked details such as what to do if a_3=0.)

    Let y = a_3x+b_3. Then x = \frac{y-b_3}{a_3}, and a_ix+b_i = \frac{a_i}{a_3}y + \frac{a_3b_i-a_ib_3}{a_3} = \alpha_iy+\beta_i say.

    The equation then becomes (\alpha_1y+\beta_1)^n + (\alpha_2y+\beta_2)^n = y^n. Compare the constant term and the coefficient of y on both sides to get the equations \beta_1^n+\beta_2^n = 0 and \alpha_1\beta_1^{n-1} + \alpha_2\beta_2^{n-1} = 0. Deduce from these that (\beta_2\alpha_1 - \beta_1\alpha_2)\beta_1^{n-1} = 0. It follows that either \beta_1=\beta_2 = 0 or \beta_2\alpha_1 = \beta_1\alpha_2. In either case, you can deduce that \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2}, from which it easily follows that all three linear polynomials are multiples of each other.
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