Why is $\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$ never a perfect square?

For all naturals $n$ we have $$f(n+1)-f(n)=44\cdot10^{2n}+8\cdot{10^n}.$$

Hence, $$\frac{f(n+1)-f(n)}{16}=11\cdot25\cdot10^{2(n-1)}+5\cdot10^{n-1}\in\mathbb N.$$

It follows that $$f(n)\equiv f(1)\equiv8\pmod{16}$$ for all $n\in\mathbb N$.

Because every square is equivalent to $0,1,4$ or $9$ modulo $16$, it follows that $f(n)$ is never a square.


Hint:

Can you show that $\sqrt{f(n)}$ is between consecutive integers $\dfrac2310^n+\dfrac13$ and $\dfrac2310^n+\dfrac43$?