About a result in Martin Davis' 1973 article "Hilbert's Tenth Problem is Unsolvable" In Martin Davis, Hilbert's Tenth Problem is Unsolvable, The American Mathematical Monthly, Vol. 80, No. 3 (Mar., 1973), pp. 233-269 (link), the author prove the following result:
Theorem 3.1: For given $a,x,k,a>1$, the system
(I) $x^2-(a^2-1)y^2=1$
(II) $u^2-(a^2-1)v^2=1$
(III) $s^2-(b^2-1)t^2=1$
(IV) $v=ry^2$
(V) $b=1+4py=a+qu$
(VI) $s=x+cu$
(VII) $t=k+4(d-1)y$
(VIII) $y=k+e-1$
has a solution in the remaining arguments $y,u,v,s,t,b,r,p,q,c,d,e$ if and only if $x=x_{k}(a)$. Here $x_{k}(a)$ pertains to the $k^{th}$ solution of Pell's equation (I), namely $(a + \sqrt{d})^k = x_k + y_k\sqrt{d}$ where $d = a^2 - 1$.
This theorem was used to prove that the exponential function $h(n,k)=n^{k}$ is Diophantine. The proof of this last result follows at once from the proof of the following lemma:
Lemma: $m=n^{k}$ if and only if equations I-VIII and
(IX) $(x-y(a-n)-m)^2=(f-1)^2(2an-n^2-1)^2$
(X) $m+g=2an-n^2-1$
(XI) $w=n+h=k+l$
(XII) $a^2-(w^2-1)(w-1)^2z^2=1$
have a solution in the remaining arguments.
I will add the proof of this result to clarify my questions:
Proof:  Suppose I-XII hold. By XI, $w>1$. Hence $(w-1)z>0$ and so by XII $a>1$. So Theorem 3.1 applies and it follows that $x=x_{k}(a),y=y_{k}(a)$.
By IX and Lemma 2.17, $$m\equiv n^k\pmod{2an-n^2-1}$$
XI yields $k,n<w$.  By XII (using Lemma 2.4, for some $j$, $a=x_{j}(w),(w-1)z=y_{j}(w)$.
By Lemma 2.14, $j\equiv 0\pmod {w-1}$, so that $j≥w-1$. So by Lemma 2.19, $a≥w^w-1>n^k$. Now by X, $m<an-n^2-1$, and by Lemma 3.4 $n^k<2an-n^2-1$
Since $m$ and $n^k$ are congruent and both less than the modulus, they must be equal.
Conversely, suppose that $m=n^{k}$. Solutions must be found for I-XII. Choose any number $w$ such that $w>n$ and $w>k$. Set $a=x_{w-1}(w)$ so that $a>1$. By Lemma 2.14, $y_{w-1}(w)\equiv 0\pmod {w-1}$. So one can write $y_{w-1}(w)≡z(w-1)$. Thus XII is satisfied. XI can be satisfied by setting $h=w-n,I=w-k$. As before, $a>n^{k}$ so that again by Lemma 3.4, $m=n^k<2an-n^2-1$ and X can be satisfied. Setting $x=x_{k}(a),y=y_{k}(a)$, Lemma 2.17 permits one to define$ f$ such that $$x-y(a-n)-m=±(f-1)(2an-n^2-1)$$ so that IX is satisfied. Finally, I-VIII can be satisfied by Theorem 3.1.
The cited results are proved in the cited paper. It seems to me that the choice of $a=x_{w-1}(w)$ is given only to prove that $a>1$. However, I am not sure about this.
Now, my questions are related to equation XII:
(1) Is the choice of $a=x_{w-1}(w)$ must be unique
(2) Can we replace XII by a simpler Pell equation such as: $a^2-(w^2-1)z^2=1$ to avoid the dependence of the indice (in the formula $a=x_{w-1}(w)$) by $w$ and choosing any $w$ such that all the required conditions for $a$ are satisfied. In this case, one can get $a=x_{j}(w)$ for some $j>1$ independent of $w$. This idea is based on the fact that $x_{j}(w)$ is strictly increasing with respect to $w$ and fixed $j$.
 A: In the original proof of Davis, equation (XI) is used only to establish precisely that $w$ exceeds both $n$ and $k$.  As there are no other restrictions on $w$, the set of $w$ is infinite. Thus, we will prove that we can consider the simpler Pell's equation $$a^2-(w^2-1)z^2=1$$ instead of the former one.
Suppose I-XII hold. By XI, $w>1$. Hence all the solutions of the equation (XII) are $$a=x_{j}(w),z=y_{j}(w)$$ with $j≥1$. Thus, we can fix $j≥1$ and choosing $w$ such that $z>0$ and $a>1$ and these are the same required conditions in the first line of the proof of Davis. So Theorem 3.1 applies and it follows that $$x=x_{k}(a),y=y_{k}(a)$$
By IX and Lemma 2.17, $$m\equiv n^k\pmod{2an-n^2-1}$$
XI yields $k,n<w$. Since $x_{j}(w)$ is strictly increasing with respect to $w$ ($j$ fixed) and the set of $w$ is infinite, then there exist a positive integer $w_{0}$ such that $$a=x_{j}(w)>n^{k}$$ for any $w≥w_{0}$. We can choose any $w_{0}$ such that $w_{0}>n$ and $w_{0}>k$. Now by X, $m<an-n^2-1$, and by Lemma 3.4 we get $n^{k}<2an-n^2-1$. Since $m$ and $n^{k}$ are congruent and both less than the modulus, they must be equal.
Conversely, suppose that $m=n^{k}$. Solutions must be found for I-XII. Choose any number $w≥w_{0}$, then $w>n$ and $w>k$. Choose $a=x_{j}(w)$ with $j≥1$ fixed and any $w≥w_{0}$ so that $a>1$. By Lemma 2.4 $$a=x_{j}(w),z=y_{j}(w)$$
Thus XII is satisfied. XI can be satisfied by setting $$h=w-n,I=w-k$$
As before, $a=x_{j}(w)>n^{k}$ for any $w≥w_{0}$, so that again by Lemma 3.4, $m=n^{k}<2an-n^2-1$ and X can be satisfied. Setting $$x=x_{k}(a),y=y_{k}(a)$$
Lemma 2.17 permits one to define $f$ such that $$x-y(a-n)-m=±(f-1)(2an-n²-1)$$ so that IX is satisfied. Finally, I-VIII can be satisfied by Theorem 3.1.
