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?
 A: 
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}
A: Hint: $n$ digits of $6$ is equal to
$$
  \left(6\left(\frac{10^n-1}{9}\right)\right)^2 = \frac{4}{9}(10^n-1)^2
     = \frac{4}{9}(10^{2n}-2\cdot 10^n+1).$$
Apply a similar transformation to the term with $8$'s, then see if you can simplify the sum.
A: Every time starting with $66^2$ adds another 4 to the beginning and a five between the 3 and 6. Also, the number of eights is half the length of $6666...^2$. 
If this pattern continues, there are (n-1) 4's, a three, (n-1) five's, and finally a six. Because you add n 8's, the number becomes n 4's.
