The content in today's blog is based on Harold M. Stark's An Introduction to Number Theory.
For Lemma 1, αn, qn are defined here. For Lemma 2, Mn is defined here.
Lemma 1: αnqn-1 + qn-2 ≠ 0
(1) Assume that αnqn-1 + qn-2 = 0.
(2) Then, qn-1 = 0 and qn-2 = 0. [See here for proof]
(3) But qn ≥ 1 for n ≥ 1. [See here]
(4) Futher, there are no successive 0's in a row since (see here for details on how these values are derived):
q-2 = 1
q-1=0
q0=1
(5) Therefore #2 is impossible and we reject our assumption.
QED
Lemma 2: if γn = αnqn-1 + qn-2, then γn(1,α) = (1,αn)Mn
(1) From the definition of Mn (see here), we know that:
= (qn-2+αnqn-1, pn-2+αnpn-1) [See here for review on matrix products]
(2) By applying the equation α = (pn-2 + αnpn-1)/(qn-2 + αnqn-1) [See here], we get:
(qn-2+αnqn-1 , pn-2+αnpn-1) = (q
(3) Adding back γn, this gives us:
(q
QED
Lemma 3: if δ = γn+j/γn, then δ(1,√d) = (1,√d)Mn-1Mn+kj
(1) From Lemma 2, we know that;
γn(1 , α) = (1, αn)Mn
and
γn+j(1, α) = (1, αn+j)Mn+j
(2) By the definition of matrix inverses (see here), we know that:
(1,αn)MnMn-1 = (1,αn)
(3) From #1, then we have:
[(1,αn)Mn)] = γn(1,α)
(4) So, from #2, we have:
(1,αn) = γn(1,α)Mn-1
(5) Likewise since αn = αn+j (see here), we get:
(1,αn+j)Mn+j = (1,αn)Mn+j
(6) Applying #4, this gives us:
(1,αn+j)Mn+j = γn(1,α)Mn-1Mn+j
(7) Combining #1 with #6, gives us:
γn+j(1,α) = γn(1,α)Mn-1Mn+j
(8) Now, δ = γn+j/γn [We can do this since γn ≠ 0 (see Lemma 1 above) ]
(9) δ(1,α) = [(γn+j)(1 , α)]/γn
(10) Applying #7 gives us;
δ(1,α) = [γn(1,α)Mn-1Mn+j]/γn =
= (1,α)Mn-1Mn+j
QED
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