# Various proofs I just want to verify

#### chaotixmonjuish

If a|(b+c) and gcd(b,c)=1, then gcd(a,b)=1=gcd(a,c)

Suppose a|(b+c) and the gcd(a,b)=d.

al=b+c and d|a and d|b. This implies a=dr and b=ds.

drl=ds+c => drl-ds=c => d(rl-s)=c => d|c

Since d|b, d|c and 1|b and 1|c, d must divide 1. Therefore d=1.

By the same reasoning gcd(a,c)=1.

al=b+c and d|a and d|c. This imples a=dr and c=dt

drl=b+dt => drl-dt=b => d(rl-t)=b => d|b

since gcd(b,c)=1 1|b and 1|c and d|c d must divide 1

therefore d=1

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#### de_brook

If a|(b+c) and gcd(b,c)=1, then gcd(a,b)=1=gcd(a,c)

Suppose a|(b+c) and the gcd(a,b)=d.

al=b+c and d|a and d|b. This implies a=dr and b=ds.

drl=ds+c => drl-ds=c => d(rl-s)=c => d|c

Since d|b, d|c and 1|b and 1|c, d must divide 1. Therefore d=1.

By the same reasoning gcd(a,c)=1.

al=b+c and d|a and d|c. This imples a=dr and c=dt

drl=b+dt => drl-dt=b => d(rl-t)=b => d|b

since gcd(b,c)=1 1|b and 1|c and d|c d must divide 1

therefore d=1
The proof is right. but note that there could be more than one integer dividing a particular number, so if 1 divides b and c it does not imply gcd(b,c) = 1. The way you have done the proof is very correct, but always note the statement
"Since d|b, d|c and 1|b and 1|c, d must divide 1."

#### chaotixmonjuish

I'm really not sure how to concluide proofs like this. How should I have concluded it?

#### de_brook

you will just rewrite he statement "Since d|b, d|c and 1|b and 1|c, d must divide 1." as follows "Since gcd(b,c) = 1 and d|b, d|c, then gcd(b, c) = 1." then u can conclude that
d = 1. hence gcd(a,b)=1=gcd(a,c)

#### chaotixmonjuish

I don't know why, but any gcd type proof always throws me off.

#### chaotixmonjuish

Ok, so
The conclusion of my proof just for gcd(a,b) should say:

Since gcd(b,c)=1 and d|b and d|c, then the gcd(a,b)=1.

#### de_brook

yes, you got it. gcd type proof is interesting

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