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Thread: Divisible by 10^9 please help!

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    Divisible by 10^9 please help!

    Let be a polynomial with real coefficients. Show that there exists a nonzero polynomial with real coefficients such that has terms that are all of a degree divisible by .
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  2. #2
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    Quote Originally Posted by usagi_killer View Post
    Let be a polynomial with real coefficients. Show that there exists a nonzero polynomial with real coefficients such that has terms that are all of a degree divisible by .
    Here's one way to approach this problem. To get an idea of what is involved, start with a simpler problem.

    Let $\displaystyle \color{blue}P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $\displaystyle \color{blue}Q(x)$ with real coefficients such that $\displaystyle \color{blue}P(x)Q(x)$ has terms that are all of a degree divisible by 2.

    A bit of thought shows that this is not too hard. Split P(x) into terms of even degree and terms of odd degree. Then we can write $\displaystyle P(x) = A(x) + xB(x)$, where both A(x) and B(x) have all terms of even degree. If $\displaystyle Q(x) = A(x)- xB(x)$ then $\displaystyle P(x)Q(x) = (A(x) + xB(x))(A(x) - xB(x)) = (A(x))^2 - x^2(B(x))^2$, which has all terms of even degree, as required. For convenience later on, define $\displaystyle P_2(P(x)) = P(x)Q(x)$.

    Now try something a bit harder.

    Let $\displaystyle \color{blue}P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $\displaystyle \color{blue}Q(x)$ with real coefficients such that $\displaystyle \color{blue}P(x)Q(x)$ has terms that are all of a degree divisible by 3.

    Let $\displaystyle \omega = e^{2\pi i/3}$, a complex cube root of unity. Check that if $\displaystyle a+bx+cx^2$ is a real quadratic polynomial then $\displaystyle (a+bx+cx^2)(a+\omega bx+\omega^2 cx^2)(a+\omega^2 bx+\omega cx^2)$ has terms that are all of a degree divisible by 3. Also, the second and third of those three factors are complex conjugates of each other, so their product is a real polynomial.

    For a general polynomial P(x), we can write $\displaystyle P(x) = A(x) + xB(x) + x^2C(x)$, where each of A(x), B(x) and C(x) has terms that are all of a degree divisible by 3. Let $\displaystyle Q(x) = (A(x)+\omega xB(x)+\omega^2 x^2C(x))(A(x)+\omega^2 xB(x)+\omega x^2C(x))$. Then $\displaystyle P_3(P(x)) = P(x)Q(x)$ is a multiple of P(x) whose terms are all of a degree divisible by 3.

    The next part of the project is to do the same thing with 3 replaced by 5. You obviously have to start with a fifth complex root of unity, $\displaystyle \omega=e^{2\pi i/5}$, and the argument gets a bit tedious, so I'll skip it. The end result is a real polynomial Q(x) (a product of four complex factors) such that $\displaystyle P_5(P(x)) = P(x)Q(x)$ is a multiple of P(x) whose terms are all of a degree divisible by 5.

    Now you can start to stitch things together. Notice that $\displaystyle P_{10}(x) = P_5(P_2(P(x)))$ is a multiple of P(x) whose terms are all of a degree divisible by 10. Then inductively define $\displaystyle P_{10^{k+1}}(P(x)) = P_{10}(P_{10^k}(P(x)))$. So finally, $\displaystyle P_{10^9}(P(x))$ is a multiple of P(x) whose terms are all of a degree divisible by $\displaystyle 10^9$.
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  3. #3
    Senior Member Shanks's Avatar
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    Wow! Amazing, A pertect application of the unit roots!
    How can you image this stuff?
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