let $x$ be a number with n digits which are all 1's. Then $A=4x(10^n+1)$, $B=8x$ so \begin{eqnarray*} A+2B+4 &=& 4x(10^n+1)+16x+4 \\ \ &=& 4x(10^n-1)+8x +16x+4 \\ \ &=& 4x\cdot 9x +24x +4 \\ \ &=& (6x+2)^2. \end{eqnarray*}

peeta wrote: let $x$ be a number with n digits which are all 1's. then $A=4x(10^n+1)$ $B=8x$ so $A+2B+4 = 4x(10^n+1)+16x+4 =4x(10^n-1)+8x +16x+4 = 4x\cdot 9x +24x +4= (6x+2)^2$ Could you please explain how you got $A=4x(10^n+1)$? Thanks.

$A = 4\cdots44\cdots4=10^n\cdot 4\cdots4 +4\cdots4=4x(10^n+1)$

You can represent those numbers as $A=444....444=4 \cdot\frac{999...999}{9}=4\cdot\frac{100..000-1}{9}=4\cdot\frac{10^{2n}-1}{9}$ $B=888...888=8 \cdot\frac{999...999}{9}=8\cdot\frac{100..000-1}{9}=8\cdot\frac{10^n-1}{9}$ $A+2B+4=4\cdot\frac{10^{2n}-1}{9}+16\cdot\frac{10^n-1}{9}+4=\frac{4 \cdot10^{2n}-4+16\cdot10^n-16+36}{9}=\frac{4 \cdot10^{2n}+16+16\cdot10^n}{9}=\frac{4(10^{2n}+4\cdot10^n+4)}{9}=(\frac{2(10^n+2)}{3})^2$ This will always be natural, because $10^n+2=100....002$ and it is always divisible with $3$

We have $A=4+4 \cdot 10 + 4 \cdot 10^2 + \ldots + 4 \cdot 10^{2n} = \frac{4(10^{2n}-1)}{9}$ and $B=8+8 \cdot 10 + 8 \cdot 10^2 + 8 \cdot 10^n = \frac{8(10^n-1)}{9}$. Hence $2B=\frac{16(10^n-1)}{9}$ and $A+2B=\frac{4 \cdot 10^{2n} + 16 \cdot 10^n - 20}{9}$, so $A+2B+4=\frac{4 \cdot 10^{2n} + 16 \cdot 10^n + 16}{9}=\left(\frac{2(10^n+2)}{3}\right)^2$. This is an integer, since $10^n + 2 \equiv 1^n + 2 \equiv 0 \pmod 3$.