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Math Help - Number Theory-GCD and Divisibility URGENT

  1. #1
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    Number Theory-GCD and Divisibility URGENT

    1. If there exists integers x and y for which ax+by=gcd(a,b), then gcd(x,y)=1.

    2. Given an odd integer a, establish that a^2+(a+2)^2+(a+4)^2+1 is divisible by 2

    3. Prove that for a positive integer n and any integer a, gcd(a, a+n) divides n; hence gcd(a, a+1)=1.
    Last edited by porterhv; September 8th 2008 at 09:01 AM.
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  2. #2
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    Hello, porterhv!

    #2 is quite simple . . .


    2. Given an odd integer a
    . . establish that a^2+(a+2)^2+(a+4)^2+1 is divisible by 2

    Since a is an odd integer, let a \:=\:2n+1 for some integer n.

    Then we have: . (2n+1)^2 + (2n+3)^2 + (2n+5)^2 + 1

    . . . . . . . . . = \;(4n^2 + 4n + 1) + (4n^2 + 12n + 9) + (4n^2 + 20n + 25) + 1

    . . . . . . . . . = \;12n^2 + 36n + 36

    . . . . . . . . . = \;12(n^2 + 3n + 3)

    The number is not only divisible by 2 . . . but also 3, 4, 6, and 12.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Acutally, I "eyeballed" the answer.


    Since \text{(odd)} + \text{(even)} \:=\:\text{odd}

    . . we have: . \text{(odd)}^2 + \text{(odd)}^2 + \text{(odd)}^2 + \text{(odd)}^2


    Since \text{(odd)}^2 \:=\:\text{(odd)}

    . . we have: . \underbrace{\text{(odd)} + \text{(odd)}} + \underbrace{\text{(odd)} + \text{(odd)}}

    . . . . . . . . . . . \underbrace{\text{(even)}\qquad + \qquad \text{(even)}}

    . . . . . . . . . . . . . . . . . \text{(even)}

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  3. #3
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    Quote Originally Posted by porterhv View Post
    1. If there exists integers x and y for which ax+by=gcd(a,b), then gcd(x,y)=1.
    Let d=\gcd(a,b) now write ax+by=d \implies (a/d)x+(b/d)y=1. Thus, \gcd(x,y)=1.

    3. Prove that for a positive integer n and any integer a, gcd(a, a+n) divides n; hence gcd(a, a+1)=1.
    Let d=\gcd(a,a+n). We know d|a and d|(a+n) thus d|(a+n)-a] \implies d|n.
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