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 !

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- Dec 12th 2009, 11:40 PM #1

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- Dec 13th 2009, 12:53 PM #2

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- Dec 13th 2009, 06:48 PM #3
Notice that gcd(n,m)=1 iff the m-class is in the $\displaystyle Z_n^*$ !

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 $\displaystyle Z_{a+b}^*$ group, so the product of a-class and b-class lies in the group, that is equivalent to say gcd(a+b,ab)=1.

- Dec 13th 2009, 11:30 PM #4

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