How to decrypt Rabin message when p=q and you have the roots from Tonelli-Shanks

Question

I've been trying to decrypt a message encrypted with Rabin as a challenge. I'm very new to cryptography and these challenges so much of this is new.

I've factorised n and determined that p=q which I understand to be incredibly weak. After many hours I discovered that the proposed decryption using extended gcd will not work.

A similar post on this site suggested using Tonelli Shanks to get the roots, which I have done. It then went on to say to use Hensel to then get the original values which will then help me decode the message, however I am unfamiliar with how to implement this and so far have not been able to get a sensible solution so would like to ask for help.

My values are:

n = 64703986196590532550677581867968606868573389071252692910980134129544137251401009133960328088692271842214498048655106618080254509684622363068406743573918979874641476333101257493419006081088753833559346504226066744706781644205324359031963711461737816475092631177676839385116576945754784715871099567521310291121
c = 60176314581676071043291067556352196274798660837188399828657574988742539250919925123769575021091715252070984470036260674221672743791229186519807702970426856963367776191049481817101068301897617186674960557150362771617316082251276579987076557148986523477838971190589062577795308359830070072697195634741564991953
p = q = 8043878802952623586394638108236704902850439411184561583961128617599719871469109041598304494567727280429349828456316270041563810531926784203271836896365511
root0 = 2187931274452861858404184425736861076518005991476611501855956036160679792394841793895180158176546375577356726244165298846056538405976359097397665134536364 
root1 = 5855947528499761727990453682499843826332433419707950082105172581439040079074267247703124336391180904851993102212150971195507272125950425105874171761829147

Answer 1

We are given $n>4$ and ciphertext $c\in(0,n)$ for textbook Rabin encryption. We want to solve for $x\in[0,n)$ the equation $x^2\bmod n=c$. We found that $n$ is a square, computed $p=\sqrt n$, found that it is prime, solved $y^2\bmod p=c\bmod p$ yielding two roots $y_0\in(0,p/2)$ and $y_1=p-y_0$, and now want the solutions for the original equation.

Every $x\in[0,n)$ can be uniquely written as $x=y+z\,p$ with $y\in[0,p)$ and $z\in[0,p)$.
If $x=y+z\,p$ is a solution of $x^2\equiv c\pmod{p^2}$, then that also yield modulo $p$ since $p$ divides $p^2$, thus $y^2\equiv c\pmod p$, thus our earlier work did yield the possible values of $y$.

Now $(y+z\,p)^2\equiv c\pmod{p^2}$ becomes $y^2+2\,y\,z\,p\equiv c\pmod{p^2}$, that is
$2\,y\,z\,p\equiv c-y^2\pmod{p^2}$, that is
$\exists k\in\mathbb Z,\,2\,y\,z\,p+k\,p^2=c-y^2$, that is
$\exists k\in\mathbb Z,\,2\,y\,z+k\,p=(c-y^2)/p$, that is
$c-y^2$ is divisible by $p$ and $2\,y\,z\equiv(c-y^2)/p\pmod p$, that is
$c-y^2$ is divisible by $p$ and $z\equiv(2\,y)^{-1}\,((c-y^2)/p)\pmod p$

Thus we check that $c-{y_0}^2\bmod p=0$ (which must hold unless we goofed in computing $y_0$),
compute $b_0=(c-{y_0}^2)/p\bmod p$, then $z_0=(2\,y_0)^{-1}\,b_0\bmod p$,
then $x_0=y_0+z_0\,p$, and that's one solution. It can't be zero. The other is $x_1=p^2-x_0$.