Let a >= b > 0 be two integers.
Let r(0)=a, r(1)=b, c(0)=0, c(1)=1, d(0)=1, d(1)=0 and

r(i) = r(i-2) -q(i-1)r(i-1) for i=2, ..., n. r(n+1)=0
c(i) = c(i-2) -q(i-1)c(i-1)
d(i) = d(i-2) -q(i-1)d(i-1)

Lemma 1. a*d(i) + b*c(i) = r(i) for i>= 0.

Lemma 2. For i>1
(1) c(i-1)c(i)<0, that is, the sign alternates.
(2) |c(i-1)| <= |c(i)|, the absolute value is non-decreasing.
Similar for d(i).

Proof. Induction. Note q(i)>=1 for all i.
Basis. i=2
(1) c(1)=1, c(2)=-a/b
(2) |c(1)|=1, |c(2)| = a/b >=1 = |c(1)|
Induction step.
(1) c(i+1)=c(i-1)-q(i)c(i) 
    case 1. c(i-1)<0, c(i)>0 ==> c(i+1)<0
    case 2. c(i-1)>0, c(i)<0 ==> c(i+1)>0
(2) |c(i+1)| = |c(i-1)-q(i)c(i)| = |c(i-1)| + |q(i)c(i)| >= |c(i)|

Lemma 3. |c(n+1)|=a, |d(n+1)|=b, if gcd(a,b)=1.
Proof. From lemma 1, a*d(n+1) + b*c(n+1) = r(n+1) = 0.
That is, a*d(n+1) = -b*c(n+1). Thus c(n+1)=-a, d(n+1)=b, or
c(n+1)=a, d(n+1)=-b, since there is no common factor in a and b.

Note. If gcd(a,b)>1, let a'=a/gcd(a,b) and b'=b/gcd(a,b).
Then lemma 3 holds for a' and b', and a'<=a and b'<=b.
 
Theorem. |c(i)| <= a, |d(i)| <= b
Proof. |c(i)| and |d(i)| are monotone non-decreasing (lemma 2) and
bounded by a and b (lemma 3).