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Math Help - Show (4n^2 + 4n + 1)^m is odd

  1. #1
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    Show (4n^2 + 4n + 1)^m is odd

    Is it possible to show (4n^2 + 4n + 1)^m is odd?

    I got to this stage in working out a question (i am not sure this stage is right in the first place)
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  2. #2
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    Quote Originally Posted by sirellwood View Post
    Is it possible to show (4n^2 + 4n + 1)^m is odd?

    I got to this stage in working out a question (i am not sure this stage is right in the first place)
    Factor 4n^2+4n+1. Then use exponent rules to simplify. The result will show that it is always odd.
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    sorry i dont know what u mean? could you talk me thought it? thanks :-)
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    Quote Originally Posted by sirellwood View Post
    sorry i dont know what u mean? could you talk me thought it? thanks :-)
    (2n+1)^2=4n^2+4n+1 A power raised to another power results in multiplying the two so the final simplification is (2n+1)^(2m). For any n, 2n+1 is odd. Odd numbers times odd numbers are always odd, no matter how many, so for any m the result is still odd.
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    Thanks. that makes perfect sense. Is it ok in a proof to just state that "odd numbers times odd numbers are always odd" or should i expand this a bit further to explain it?
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    Quote Originally Posted by sirellwood View Post
    Thanks. that makes perfect sense. Is it ok in a proof to just state that "odd numbers times odd numbers are always odd" or should i expand this a bit further to explain it?
    I think it's simple enough to not need proof but if you want to it's not hard.

    Take a list of odd numbers multiplied: O_1*O_2...O_n. If this were even then that product would divide by 2 with no remainder. Since no individual odd can be divided by 2, no O_i in the list will divide by it and no O_i can be a multiple of 2, by definition of odd.
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