i just want to ask if why does the modulo affect the other terms in an eqution? Why does the 4th equation has to be multiplied by $a^2$? Then as the modulo becomes $n≡1(mod3)$ in the 5th eq. then the power of 10 in the second term becomes 5?Ive been thinking for reasons why it happens but i cant find any, please help. Im currently studyng the journal about POWERS
\begin{equation}
\label{eq:(3)} {\overline{aa \ldots ab \ldots b}}_{(10)} = y^2,
\end{equation} $$
10^{m} \cdot a \frac{10^{n} - 1}{9} = y^2 ,
$$
$\bullet$ If $m = 3$, then equation 1 yields
$$
10^3 aa \ldots a + 111b = y^2.
$$
Hence,
\begin{equation}
\label{eq:(4)} 10^{n+3} a - 10^3 a + 999b = (3y)^2.
\end{equation}
If $n\equiv 0\pmod 3$; i.e., if $n = 3N$ for a some integer $N$,
then equation 1 yields
\begin{equation}
\label{eq:(5)} Y^2 = X^3 - 10^3 a^3 + 999 a^2 b,
\end{equation}
where $Y = 3ay$ and $X = 10^{N+1}a.$
If $n \equiv 1\pmod 3$; i.e., if $n = 3N + 1$ for some integer $N$, then equation 1 yields \begin{equation} \label{eq:(6)} Y^2 = X^3 - 10^5 a^3 + 99900 a^2 b, \end{equation} where $Y = 30ay$ and $X = 10^{N+2}a.$ If $n \equiv 2\pmod 3$; i.e., if $n = 3N + 2$ for some integer $N$, then equation 1 yields \begin{equation} \label{eq:(7)} Y^2 = X^3 - 10^7 a^3 + 9990000 a^2 b, \end{equation} where $Y = 300ay$ and $X = 10^{N+3}a.$
You start with $$10^{n+3}a-10^3a+999b=(3y)^2$$ You let $n=3N+1$ to get $$10^{3N+4}a-10^3a+999b=(3y)^2$$ You multiply by 100a^2 to get $$10^{3N+6}a^3-10^5a^3+99900a^2b=(30ay)^2$$ Then you let $Y=30ay$, and $X=10^{N+2}a$ to get $$X^3-10^5a^3+99900a^2b=Y^2$$