444...44+2×888...88+4 is perfect square?

Question

Prove that $\underbrace{444\dots44}_{2n} + 2×\underbrace{888\dots88}_{n} + 4$ is a perfect square

For me I dunno how to do this, I could only managed to notice that 4 is perfect square. Please help

Answer 1

As pointed out in the comments, the proposition as stated is not true, but we can say this:

$$\underbrace{666...666}_{n-1}7^2=\left(6\times\dfrac{10^n-1}9+1\right)^2=\left(\dfrac{6\times10^n+3}{9}\right)^2=\left(\dfrac {2\times10^n+1}{3}\right)^2$$

$$=\dfrac{4\times10^{2n}+4\times10^n+1}9=4\times\dfrac{10^{2n}-1}9+4\times\dfrac{10^n-1}9+1=\underbrace{444...444}_n\underbrace{888...888}_{n-1}9$$

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Addendum in response to editing of question:

My answer above was a reponse to the original question,

which asked to prove that $\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$ is a perfect square, which it never is.

Now that the question asks to prove that $\underbrace{444\dots44}_{2n} + \color{blue}{2\times}\underbrace{888\dots88}_{n} + 4$ is a perfect square,

note that $m^2=\underbrace{444\dots44}_{2n} + \color{black}{2\times}\underbrace{888\dots88}_{n} + 4=4\times\left(\dfrac{10^{2n}-1}9\right)+2\times8\times\dfrac{10^n-1}9+4$

$\implies\dfrac94m^2=(10^{2n}-1)+4(10^n-1)+9= 10^{2n}+4\times10^n+4=(10^{n}+2)^2$

$\implies\dfrac32m=10^{n}+2\implies m=\dfrac23(10^n+2),$

and $10^n+2$ is divisible by $3$ because $10$ leaves remainder $1$ when divided by $3,$ so $m$ is an integer.