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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