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Math Help - Divisible by 16

  1. #1
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    Divisible by 16

    Prove that if m and n are odd integers, then m^3n^2+m^2n^3-m-n is divisible by 16.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sigyn3 View Post
    Prove that if m and n are odd integers, then m^3n^2+m^2n^3-m-n is divisible by 16.
    this seems like a straight forward computation will work

    let m = 2a + 1 and let n = 2b + 1

    where a and b are integers

    expand the expression and simplify. your goal is to show that the expression can be written as 16k where k is an integer
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  3. #3
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    Hello, Sigyn3!

    Prove that, if m and n are odd integers, then m^3n^2+m^2n^3-m-n is divisible by 16.
    I used your approach, Jhevon . . . and hit a wall.


    We have: . P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)


    Let . \begin{array}{ccc}m &=&2a+1 \\ n &=&2b+1 \end{array}

    Then:. . \begin{array}{ccccccc}m + n &=& (2a+1) + (2b+1) &=& \quad 2a+2b + 2 &\;\;=& \quad2(a+b+1)\\<br />
mn-1 &=& (2a+1)(2b+1) - 1 &=& \quad 4ab + 2a +2b &\;\;=& \;\;2(2ab + a + b) \end{array}
    . . . . . . \begin{array}{ccccccc}mn+1 &=& (2a+1)(2b+1)+1 &=& 4ab + 2a + 2b + 2 &=& 2(2ab + a + b + 1)\end{array}


    And we have: . P \;=\;2(a + b + 1)\cdot2(2ab + a + b)\cdot2(2ab + a + b + 1)

    . . . . . . . . . . P \;=\;8(a+b+1)(2ab + a + b)(2ab + a + b + 1)

    So we have a multiple of 8.
    . . And here's where I hit the wall . . .


    I tried various ways to find another factor of 2,
    . . and settled on an exhaustive listing.

    . . \begin{array}{c|c|c|c}<br />
(a,b) & (a+b+1) & (2ab + a + b) & (2ab + a + b + 1) \\ \hline<br />
\text{even, even} & \text{odd} & \text{even} & \text{odd} \\ <br />
\text{even, odd} & \text{even} & \text{odd} & \text{even} \\<br />
\text{odd, odd} & \text{odd} & \text{even} & \text{odd} \end{array}

    So for any combination of a and b, at least one of the factors is even.

    Therefore, P is a multiple of 16.



    Surely there must a more elegant method . . .
    .
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  4. #4
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    Soroban,
    Whenever you said:
    <br />
P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)
    you've already solved it.

    Because:
    m+n is even.
    mn-1 is even
    mn+1 is even (AND next even number after mn-1)
    which means either mn-1 or mn+1 is divisible by 4.
    <br />
(m+n)(mn-1)(mn+1) = 2k_1 2k_2 4k_3 = 16k_4
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Sigyn3 View Post
    Prove that if m and n are odd integers, then m^3n^2+m^2n^3-m-n is divisible by 16.
    First note that the square of any odd integer is congruent to 1 modulo 8. This is because (2k+1)^2=4k(k+1)+1 and one of k and k+1 is even.

    Since m snd n are odd, so is mn. \therefore\ m^2n^2\equiv1\pmod8

    \Rightarrow\ m^2n^2-1 is divisible by 8.

    And since m+n is even, m^3n^2+n^3m^2-m-n=(m+n)(m^2n^2-1) is divisible by 16.
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