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Thread: 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; Sep 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 $\displaystyle a$
    . . establish that $\displaystyle a^2+(a+2)^2+(a+4)^2+1$ is divisible by 2

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

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

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

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

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

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


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


    Acutally, I "eyeballed" the answer.


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

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


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

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

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

    . . . . . . . . . . . . . . . . . $\displaystyle \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 $\displaystyle d=\gcd(a,b)$ now write $\displaystyle ax+by=d \implies (a/d)x+(b/d)y=1$. Thus, $\displaystyle \gcd(x,y)=1$.

    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 $\displaystyle d=\gcd(a,a+n)$. We know $\displaystyle d|a$ and $\displaystyle d|(a+n)$ thus $\displaystyle d|(a+n)-a] \implies d|n$.
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