Length of Hirzebruch continued fractions Suppose $a,b$ are two natural numbers relatively prime to $n$ and to each other. Assume $n\geq ab+1$. Suppose further that $\frac{a}{b}\equiv k \pmod{n}$ for some $k\in \lbrace 1,2,\dots, n-1\rbrace$ and $\frac{a}{b}\equiv k'\pmod{n+ab}$ for some $k'\in \lbrace 1,2,\dots, n+ab-1\rbrace$.

Question: Is there an elementary proof that the length of the continued fraction of $\frac{n}{k}$ is equal to the length of the continued fraction of $\frac{n+ab}{k'}$?

This came out of a broader result, and for this particular case I can prove it using routine toric geometry, however I would like to know of some elementary tricks to deal with continued fractions. 

Here by continued fraction I mean the Hirzebruch continued fraction 
$$\frac{n}{k}=a_0-\frac{1}{a_1-\frac{1}{a_2-\cdots}}.$$
For example, when $a=2, b=3$ and $n=17$, we get $k=12$ and $k'=16$, so the fractions are 
$$\frac{17}{12}=2-\frac{1}{2-\frac{1}{4-\frac{1}{2}}}\qquad and \qquad\frac{23}{16}=2-\frac{1}{2-\frac{1}{5-\frac{1}{2}}}.$$
 A: Forgive me, this should be in the comments, but I am still building my reputation up to comment. If a=5, b=4, n=7, then k=3, k'=8, and n+ab=27. 
Here, $\frac{n}{k}=\frac{7}{3}=3-\frac{1}{2-\frac{1}{2}}$ but $\frac{n+ab}{k'}=\frac{27}{8}=4-\frac{1}{2-\frac{1}{3-\frac{1}{2}}}$. Am I missing something or is there a further assumption on these numbers?
A: Lets call expansions
$$\langle
x_1,\ldots,x_m\rangle:=\cfrac{1}{x_1-{\atop\ddots\,\displaystyle{-\cfrac{1}{x_m}}}}$$
(as in Perron's book) reduced regular continued fractions (RRCF).
Probably they are older then Hirzebruch.
We'll prove more precise statement.
Theorem. If $(n,ab)=1$ and $n>ab$ then RRCF for all numbers
$$\left\{\frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\}\qquad(k\ge
0)$$ are almost equal: they have equal length and differ only in
one partial quotient.
Remark 1. Common factors of $a$ and $b$ can be moved into $k$.
If $d=(a,b)$, $a=da_1$, $b=db_1$, then \begin{gather*} \left\{
\frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\} =\left\{
\frac{a_1b_1^{-1}\pmod{(n+kab)}}{n+kab}\right\} \\=\left\{
\frac{a_1b_1^{-1}\pmod{(n+(kd^2)a_1b_1)}}{n+(kd^2)a_1b_1}\right\}.
\end{gather*}
So we can assume that $(a,b)=1$.
Remark 2. The proof will be given in terms of modified
continuants $K(x_1,\ldots, x_n)$ (see ``Concrete Mathematics'' for
more explanations). These polynomials are defined by initial
conditions
$$K()=1,\quad K(x_1)=x_1$$
and the following recurrence:
$$K(x_1,\ldots,
x_n)=x_nK(x_1,\ldots, x_{n-1})-K(x_1,\ldots,
x_{n-2})\qquad(n\ge2).$$ (In the usual definition minus must be
replaced by plus.) For convenience $K_{-1}:=0$ (empty RRCF is
$0$).
In terms of continuants RRCF can be written as
$$\langle x_1,\ldots,x_n\rangle=\frac{K(x_2,\ldots,
x_n)}{K(x_1,\ldots, x_n)}.$$
Continuant's properties. All these properties can be proved by
induction (or from ``Euler’s rule'').
1$^{\circ}.$ $K(x_1,\ldots, x_n)=K(x_n,\ldots, x_{1})$.
2$^{\circ}.$ \begin{gather*} K(x_1,\ldots, x_n,x_{n+1}, \ldots,
x_{m+n})\\=K(x_1,\ldots, x_n)K(x_{n+1}, \ldots,
x_{m+n})-K(x_1,\ldots, x_{n-1})K(x_{n+2}, \ldots, x_{m+n})
\end{gather*}
3$^{\circ}.$ $\begin{vmatrix}
     K(x_2,\ldots, x_{n-1})&K(x_2,\ldots, x_n) \\
     K(x_1,\ldots, x_{n-1})&K(x_1,\ldots, x_n)
\end{vmatrix}=-1$. In particular if
$$\frac{A}{a}=\left<r_1, \ldots, r_v\right>=\frac{K(r_2, \ldots,
r_v)}{K(r_1, \ldots, r_v)}$$ then
$$K(r_1, \ldots, r_{v-1})=A^{-1}\pmod{a},\qquad K(r_2, \ldots, r_{v-1})=\frac{AA^{-1}\pmod{a}-1}{a}.$$
4$^{\circ}.$ Euler's identity (see A Short Proof of Euler's Identity
for Continuants for additional arguments). (2$^{\circ}$ and
3$^{\circ}$ are special cases of this identity) $$
K(x_1, \ldots, x_{m+n})K(x_{m+1}, \ldots, x_{m+l})-K(x_1, \ldots,
x_{m+l})K(x_{m+1}, \ldots, x_{m+n})$$ $$+K(x_1, \ldots,
x_{m-1})K(x_{m+l+2}, \ldots, x_{m+n})=0.
$$
Proof of the Theorem. For a given $n$ define
$a^{-1}:=a^{-1}\pmod{n}$, $b^{-1}:=b^{-1}\pmod{n}$, $0\le a,b\le n-1$
(inverse number is always least possible nonnegative) and $t_a$,
$t_b$ such that $aa^{-1}=1+t_an$, $bb^{-1}=1+t_bn$. Let $$
\frac{A}{a}=\left\{\frac{bt_a}{a}\right\}=\left<r_1, \ldots,
r_v\right>,\qquad A^{-1}:=A^{-1}\pmod{a};$$
$$\frac{B}{b}=\left\{\frac{at_b}{b}\right\}=\left<q_1, \ldots,
q_u\right>,\qquad B^{-1}:=B^{-1}\pmod{b};$$
$$\frac{P(x)}{Q(x)}=\left<q_1, \ldots, q_u,x,r_v, \ldots,
r_1\right>.
$$
By 2$^{\circ}$, 3$^{\circ}$ and main recurrence $$
Q(x)=K(q_1, \ldots, q_u,x,r_v, \ldots, r_1)=$$
$$xK(q_1, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_1, \ldots,
q_{u-1})K(r_v, \ldots, r_1)-K(q_1, \ldots, q_u)K(r_{v-1}, \ldots,
r_1)$$ $$=xab-aB^{-1}-bA^{-1}.
$$
Hence $$ Q(x)\equiv -bA^{-1}\equiv -b((bt_a)^{-1}\pmod{a})\equiv
-t_a^{-1}\equiv n\pmod{a},$$ $$Q(x)\equiv -aB^{-1}\equiv
-a((at_b)^{-1}\pmod{b})\equiv -t_b^{-1}\equiv n\pmod{b}.$$
Therefore
$$Q(x)\equiv n\pmod{ab},$$
and for some integer $x_0$ we have $Q(x_0)=n$. We know that $n>ab$. It
means that
$$x_0ab-aB^{-1}-bA^{-1}>ab,$$
so $x_0\ge 2$ and $\left<q_1, \ldots, q_u,x_0,r_v, \ldots,
r_1\right>$ is really RRCF. Choosing arbitrary $x=x_0+k$ we'll get
progression $n+kab$ as in the statement of the theorem.
Let's check the numerator $P(x)$. Final step $P(x)\equiv
ab^{-1}\pmod{Q(x)}$ follows from identity
$$bP(x)-BQ(x)=a,$$
which is a special case of Euler's identity. Nevertheless this
identity can be verified directly with help of
1$^{\circ}$--3$^{\circ}$: $$
P(x)=K(q_2, \ldots, q_u,x,r_v, \ldots, r_1)=$$
$$xK(q_2, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_2, \ldots,
q_{u-1})K(r_v, \ldots, r_1)-K(q_2, \ldots, q_u)K(r_{v-1}, \ldots,
r_1)$$ $$=xaB-a\frac{BB^{-1}-1}{b}-BA^{-1},$$ $$
bP(x)-a=B(xab-aB^{-1}-bA^{-1})=BQ(x).
$$
