In this question
For what values of $l$ and $j$ (simultaneously) is the following congruence solvable?
I do not understand how are $$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$$ calculated?
Also, I do not understand how is $$ xp^{s}j p^{a-b} + yp^t-l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$ this $$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$Could someone show me the details of the calculations please?
Here is the part of the answer containing those parts:
The transformation
Let $l=Lp^U$ and $j=Jp^V$ where $LJ \not\equiv 0\pmod p$. Then let $$A=a-b+s+V,B=t,C=2s+U,D=b.$$
Theorem 2
For non-negative integers $a\ge b,s$ and $t$, the equation $$l (xp^s)^2 \equiv xp^{s}j p^{a-b} + yp^t\pmod{p^b}$$ can be solved with $xy \not\equiv 0\pmod p$ if and only if the equation $$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$ can be solved with $XYZ \not\equiv 0\pmod p$.
Proof
If the $x,y$ equation can be solved then simply take $$X=xJ,Y=y,Z=-Lx^2.$$
Conversely, suppose we have a solution of the $X,Y,Z$ equation. Since $X$ and $L$ are coprime to $p$ we can solve $$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}.$$ Then $$ xp^{s}j p^{a-b} + yp^t-l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$ and this is $$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$
Q1.
$LX$ and $p^D$ are coprime. Therefore there are integers $\alpha$ and $\beta$ such that $\alpha LX+\beta p^D=1$.
Then $\alpha LX\equiv 1 \pmod {p^D}$ and so $-\alpha JZ LX\equiv -JZ \pmod {p^D}$. Then take $x=-\alpha JZ$.
Similarly for the other congruence.
Q2.
Substitute $j=Jp^V$ and $l=Lp^U$ into $xp^{s}j p^{a-b} + yp^t-l(xp^s)^2$ and you will obtain $xJp^{s+a-b+V}+yp^t-x^2Lp^{s+2U}=xJp^A+yp^B-Lx^2p^C.$
Now substitute the expressions $x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$ to obtain $$-\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C ),$$ where we know that $Xp^A+Yp^B+Zp^C\equiv 0 \pmod p^D$.