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Thread: A Problem of Polynomial

  1. #1
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    Thumbs up A Problem of Polynomial

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

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

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

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