I have a question about p-adic numbers related to Fermat's last theorem.
Introduction:
Let:
$$x^p = y^p + z^p$$ $$ \gcd(x,y,z) = 1$$ $$xyz \not \equiv 0 \pmod p$$ $$x^{p-1} \equiv y^{p-1} \equiv z^{p-1} \equiv 1 \pmod{p^3}$$
Note:
$$\frac{x^{p+1}-y^{p+1}}{x-y} = (p+1)x^p+a(x-y) \equiv x+y \pmod{p^3}$$ $$\frac{x^{p+1}-z^{p+1}}{x-z} = (p+1)x^p+b(x-z) \equiv x+z \pmod{p^3}$$ $$x \equiv y+z \pmod{p^3}$$
$\implies$
$$az \equiv y-px \pmod{p^3}$$ $$by \equiv z-px \pmod{p^3}$$
After manipulations:
$$(a+1)^p-a^p-1 \equiv (b+1)^p-b^p-1 \equiv 0 \pmod{p^3}$$
Calculating Fermat-quotients of $a,a+1$(by symmetry this can be used for $b$ as well):
$$z(a+1) \equiv y+z-px \equiv x(1-p) \pmod{p^3}$$
$\implies$
$$(a+1)^{p-1} \equiv (1-p)^{p-1} \equiv 1+p \pmod{p^3}$$
And:
$$a^{p-1} \equiv 1+p\left(\frac{a+1}{a}\right) \pmod{p^3}$$
Question:
I have seen in the text: On Congruences Related to the First Case of Fermat's Last Theorem by Wells-Johnson which is about p-adic numbers that the above leads to:
$$v(a+1) \equiv (a+1)^{p^{n+1}} \equiv (a+1)^{p^n} + r_{a+1}p^{n+1} \pmod{p^{n+2}}$$ $$v(a) \equiv a^{p^{n+1}} \equiv a^{p^n} + r_{a}p^{n+1} \pmod{p^{n+2}}$$
with $r_a = r_{a+1} = a+1$
Which should give(that is my question):
$$v(a+1)=v(a)+1$$
$\implies$
$$a^2+a+1 \equiv 0 \pmod{p}$$
with $a \equiv y/z \pmod{p}$.
This would lead to two solutions $a,b$:
$$a \equiv -\frac{p+1+(p-1)\sqrt{-3}}{2} \pmod{p^3}$$ $$b \equiv -\frac{p+1-(p-1)\sqrt{-3}}{2} \pmod{p^3}$$
and imply: $x^2+y^2+z^2 \equiv 0 \pmod{p}$ in the first case of Fermat's last theorem.
The main question is if:
$$r_a = r_{a+1} = a+1 \implies v(a+1) = v(a) + 1$$