It's better, and it's probably enough to get the marks, but it's still a bit unclear.
FWIW, while keeping your basic argument I'd probably argue something like this:
Claim: Our sequence
(a1,b1),... must eventually have a pair with
an>0,bn=0.Proof of claim: First note that since
an=bn−1 this is equivalent to the claim that
bn−1>0,bn=0 for some n. Since we start with b_1 > 0 and we have shown
b1>b2>..., the only way this could fail is if
bn>0,bn+1<0 for some n. We show this is impossible. For since
an+1=bn (and so is > 0), in this case we would have
an+1bn+1+1≤0; but then since
k=(an+12+bn+12)/(an+1bn+1+1) this contradicts our definition of k as a postiive integer.
To prove that k must be a perfect square, all that remains is to note that if
bn=0 then
k=(an2+bn2)/(anbn+1)=an2.