(s,t) is also 1.

(a,b)=1 ⇔ ∃s,t(sa+tb=1) ⇔ (s,t)=1

Let ∃s,t(sa+tb=1) and (s,t)=d≠1.

Then we get s=ds', t=dt' (s',t' in Z).

So sa+tb=dk≠1 (where k=s'a+t'b, k in Z), which is contradictory to ∃s,t(sa+tb=1).

By proof by contradiction, (i) is true.

The reverse of (i) is omitted.