1. ## Can you help me about GCD.?

I cannot proof 2 problem.....*-*!!!

1.Prove that if gcd(a,b)=1,then gcd(a+b,ab)=1.

2.If gcd(a,b)=1,and c|(a+b),then gcd(a,c)=gcd(b,c)=1.
(I can show proof gcd(a,c)=1 but I can show gcd(b,c)=1.)

....Thank !

2. Originally Posted by konna
I cannot proof 2 problem.....*-*!!!

1.Prove that if gcd(a,b)=1,then gcd(a+b,ab)=1.

Say $gcd(a+b,ab)=d\Longrightarrow$:

(1) $a+b=kd$

(2) $ab=md$

Now multiply (1) by b: $ab+b^2=(kb)d$ and substract (2): $b^2=(kb-m)d\Longrightarrow d\mid b^2\Longrightarrow d\mid b$

Now do as above but multiplying (1) by a and substracting (2) and you'll get that also $d\mid a$, so...

Tonio

2.If gcd(a,b)=1,and c|(a+b),then gcd(a,c)=gcd(b,c)=1.
(I can show proof gcd(a,c)=1 but I can show gcd(b,c)=1.)

....Thank !
.

3. Notice that gcd(n,m)=1 iff the m-class is in the $Z_n^*$ !
since gcd(a,b)=1, we have gcd(a+b,a)=1, and gcd(a+b,b)=1 by the euler algrithem.
thus gcd(a+b,ab)=1, that is because the statement "gcd(a+b,a)=1, and gcd(a+b,b)=1 " mean that a and b are in the $Z_{a+b}^*$ group, so the product of a-class and b-class lies in the group, that is equivalent to say gcd(a+b,ab)=1.

4. Thank you so much , I'm cheerful !