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Math Help - Proof that an odd number raised to an integral power is odd

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    Proof that an odd number raised to an integral power is odd

    Sorry if this is too elementary for this sub-forum.

    With n = 1, 2, 3, 4 I can see that (2k+1)^n will produce polynomials that would consist of terms with even coefficients and 1^n, which is to say odd numbers. How can I prove this for all n?
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    Quote Originally Posted by Discriminant View Post
    Sorry if this is too elementary for this sub-forum.

    With n = 1, 2, 3, 4 I can see that (2k+1)^n will produce polynomials that would consist of terms with even coefficients and 1^n, which is to say odd numbers. How can I prove this for all n?
    You can simply use that odd times odd equals odd.

    (And for completeness you might want to address the negative integers.)
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    Two ways:
    1) Use the bionomial theorem:
    (2k+1)^n= \sum_{i=0}^n \begin{pmatrix}n \\i\end{pmatrix}2^ik^i= 2\sum_{i=1}^n\begin{pmatrix}n \\i\end{pmatrix}2^{i-1}k^i+ 1

    2) Proof by induction:
    Certainly if n= 1, (2k+ 1)^1= 2k+ 1 is odd. Suppose that, for some i, [tex](2k+ 1)^i[tex] is odd. Then (2k+1)^{i+1}= (2k+1)^i(2k+1) is the product of two odd numbers and so odd, as undefined says.

    (To prove "the product of two odd number is odd" just look at (2k+ 1)(2j+1).)
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    Quote Originally Posted by Discriminant View Post
    Sorry if this is too elementary for this sub-forum.

    With n = 1, 2, 3, 4 I can see that (2k+1)^n will produce polynomials that would consist of terms with even coefficients and 1^n, which is to say odd numbers. How can I prove this for all n?

    \forall\,n\in\mathbb{N}\,,\,\,(2k+1)^n=\sum\limits  ^n_{i=0}(2k)^i , according to Newton's binom, and this is a sum of

    even numbers except for the first index i=0 , for which we have (2k)^0=1 , and thus the result is, by definition, an odd number.

    Tonio
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    Quote Originally Posted by undefined View Post
    You can simply use that odd times odd equals odd.

    (And for completeness you might want to address the negative integers.)
    (i) (2n+1)(2n+1) = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1
    (ii) (2n+1)^{-n} = \frac{1}{2n + 1}. An odd number divided by an odd number will produce an odd number, for:
    (iii) \frac{2n+1}{2n+1} = 1, which is odd.
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    Quote Originally Posted by Discriminant View Post
    (i) (2n+1)(2n+1) = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1
    (ii) (2n+1)^{-n} = \frac{1}{2n + 1}. An odd number divided by an odd number will produce an odd number, for:
    (iii) \frac{2n+1}{2n+1} = 1, which is odd.
    ??? You should recheck what you wrote

    Odd times odd is odd, we learn this in primary school, proof without modular arithmetic (2k+1)(2m+1) = 4km+2k+2m+1

    Use weak induction on exponent n in {0,1,2,...} and for negative exponents note that the result is only an integer if absolute value of base is 1.

    Edit: HallsofIvy already did it.
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    Quote Originally Posted by undefined View Post
    ??? You should recheck what you wrote

    Odd times odd is odd, we learn this in primary school, proof without modular arithmetic (2k+1)(2m+1) = 4km+2k+2m+1

    Use weak induction on exponent n in {0,1,2,...} and for negative exponents note that the result is only an integer if absolute value of base is 1.

    Edit: HallsofIvy already did it.
    Thanks! I got confused there with your previous note about completeness and the negative exponents and produced complete nonsense, which attributes parity to fractions. =)
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    Quote Originally Posted by Discriminant View Post
    Sorry if this is too elementary for this sub-forum.

    With n = 1, 2, 3, 4 I can see that (2k+1)^n will produce polynomials that would consist of terms with even coefficients and 1^n, which is to say odd numbers. How can I prove this for all n?
    I jsut want to show a different way. The result is clear for 1 because 1^n=1=2\cdot0+1 , odd. By the Fundamental Theorem of Arithmetic any integer m>1 has a unique factorizaton into primes (not considering the order), say m=p_1^{m_1}p_2^{m_2}\cdots{p_r}^{m_r} ; the exponents are positive. Then m^n=p_1^{nm_1}p_2^{nm_2}\cdots{p_r}^{nm_r} .

    It follows that n^m and n have precisely the same prime divisors. In particular, 2|m if and only if 2|m^n, i.e., if m is of the form 2k+1 for some k integer, then m^n is of the same form.
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