Unusual result to the addition 
Question: Prove that
(666... to n digits)^2 + (888... to n digits)=(444... to 2n digits)

My way: I just proved the given equation for three values of n and written at the bottom.
"Since the equation satisfies for n=1, 2, and 3, the equation is true and hence proved."
Also I am seeing a regular pattern in (6666...to n digits)^2
$666^2=443556,6666^2=44435556,66666^2=4444355556$
So the pattern is first there are (n-1) $4's$ the one $3$ then (n-1) 5's the one 6.
Is there a general way to solve this question?


Question: Prove that
$(666\dots \text{to $n$ digits})^2 + (888\dots \text{to $n$
> digits})=(444\dots \text{to $2n$ digits})$

By dividing by $4$, this is equivalent to
\begin{align}
(333\dots \text{to $n$ digits})^2 
+ (222\dots \text{to $n$ digits})
&=(111\dots \text{to $2n$ digits})
\end{align}
Let $x=111\dots \text{to $n$ digits}$. Then
\begin{align}
(3x)^2+2x&=x\cdot 10^n+x
\\
9x^2+2x&=x\cdot 10^n+x
\\
x(9x+2)&=x\cdot 10^n+x
\\
x(9x+1+1)&=x\cdot 10^n+x
\\
x(10^n+1)&=x\cdot 10^n+x.
\end{align}
