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Math Help - n^3+n=m^4 proving n has to be even and divisble by 16

  1. #1
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    n^3+n=m^4 proving n has to be even and divisble by 16

    How do you prove that n^3+n=m^4 n has to be even and also divisble by 16

    Thanks
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  2. #2
    Member anonimnystefy's Avatar
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    Re: n^3+n=m^4 proving n has to be even and divisble by 16

    hi jjy0127

    let's look at the parity of the expression n^3+n

    1. case-n is even:

    when n is even,n^3 is also even,so n^3+n is even.

    2. case-n is odd

    when n is odd,n^3 is also odd,so n^3+n is even (because a sum of two odd numbers is an even number)

    so we have that the right side is always even and divisible by 2.

    now,because the left side is divisible by 2 so must be the right side.so:
    2|m^4
    which is equivalent to:
    2|m
    so if m is divisible by two you can write it like this:
    m=2k where k is a natural number.

    from here you get that:
    m^4=(2^4) * (k^4)=16*(k^4)
    which means that m is divisible by 16

    can you continue?
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    Re: n^3+n=m^4 proving n has to be even and divisble by 16

    Thank you for your reply, it seems as though question is asking n has to be even not n^3+n is even.
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