# Number Theory-GCD and Divisibility URGENT

• Sep 8th 2008, 08:02 AM
porterhv
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.
• Sep 8th 2008, 09:21 AM
Soroban
Hello, porterhv!

#2 is quite simple . . .

Quote:

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.

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

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

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

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

• Sep 8th 2008, 10:52 AM
ThePerfectHacker
Quote:

Originally Posted by porterhv
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, $\displaystyle \gcd(x,y)=1$.

Quote:

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 $\displaystyle d|(a+n)-a] \implies d|n$.