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I How to know if a polynomial is odd or even?

  1. Jul 7, 2016 #1
    3(2k+1)3

    I have written a program which calculates the value of that polynomial with different values of k. The result is always an odd number. I am having a difficult time writing a proof that states that this polynomial always returns an odd number.

    I know that (2k + 1) is the general form for odd numbers. My polynomial does have (2k+1) in it, but it is altered.

    Here are two different polymorphs of the same polynomial. They are just simplified versions:

    3(8k3 + 12k2 + 6k + 1)

    24k3 + 36k2 + 18k + 3

    How can I prove that this is always odd?
     
  2. jcsd
  3. Jul 7, 2016 #2

    fresh_42

    Staff: Mentor

    ##3(2k+1)^3## has only odd factors. No ##2## in sight.
     
  4. Jul 7, 2016 #3
    I have the answer. 3 times any odd makes an even, but the third power of that even is always odd.
     
  5. Jul 7, 2016 #4

    fresh_42

    Staff: Mentor

    3 times odd is still odd. Even numbers can be divided by 2, odd cannot.
     
  6. Jul 7, 2016 #5
    How would you know if a difficult statement like 3(2k+1)3 can be divided by 2
     
  7. Jul 7, 2016 #6

    fresh_42

    Staff: Mentor

    It can't. ##3(2k+1)^3 = 3 \cdot (2k+1) \cdot (2k+1) \cdot (2k+1)## and every one of them is odd. There cannot be a factor ##2##. ##2## is a prime so it must divide a factor if it would divide the product. This is not the case, as long as ##k## is a natural or integer number. Of course you get an even product if you allow, e.g. ##k=1.5##.
     
  8. Jul 7, 2016 #7

    pwsnafu

    User Avatar
    Science Advisor

    If you know that the general form of odd numbers is ##2 n + 1## then all you need to do is write ##24k^3 + 36k^2 + 18k + 3## in that form, i.e. find ##n##. All you need to do is write the 3 as ##2+1## then factor out the twos.
     
  9. Jul 8, 2016 #8

    Mark44

    Staff: Mentor

    It's not necessary to multiply out the (2k + 1)3 factor.

    Assuming that k takes on only integer values,
    2k is always even, and 2k + 1 is odd.
    (2k + 1)3 is the product of three odd integers, so is itself odd.
    Muliplying an odd integer by 3 results in an odd integer, which by definition is not divisible by 2.
    All of my statements here can be made rigorous with very little effort.

    BTW, 3(2k+1)3 is an expression, not a statement. An expression has a value; a statement is either true or false. Equations and inequalities are examples of statements.
     
  10. Jul 8, 2016 #9

    Math_QED

    User Avatar
    Homework Helper

    3(2k+1)^3
    = 24k^3 + 36k^2 + 18k + 3
    = 24k^3 + 36k^2 + 18k + 2 + 1
    = 2*12k^3 + 2*18k^2 + 2*9k + 2*1 + 1
    = 2(12k^3 + 18k^2 + 9k + 1) + 1

    Then, 12k^3 + 18k^2 + 9k + 1 is a number that depends on the value of k (It doesn't matter whether this number is odd or even, since it is multiplied by 2, and 2*odd number = even number and 2*even number is even number). Now, let s = 12k^3 + 18k^2 + 9k + 1.

    Therefor, we can write 2(12k^3 + 18k^2 + 9k + 1) + 1 as 2s + 1, which is the representation of an odd number.
     
    Last edited: Jul 8, 2016
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