Why is $\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$ never a perfect square? In this question, the questioner asked to prove that $$f(n)=\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$$ is a perfect square for all $n\in\mathbb N$. However, I was not able to find any $n$ for which this is true. So I have the following "counter-conjecture": For any $n\in\mathbb N=\{1,2,3,4,\dots\}$, $f(n)$ is not a perfect square.
How can I prove this?
Notice that by simply evaluating the geometric series we get $$f(n)=\frac{8}{9} \left(10^n-1\right)+\frac{4}{9} \left(10^{2 n}-1\right)+4=\frac{4}{9} \left(2^{n+1} 5^n+100^n+6\right),$$
so it would be enough to prove that $$\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$$ is never an integer.
But I don't know how to do this.
 A: 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.
A: Hint:
Can you show that $\sqrt{f(n)}$ is between consecutive integers $\dfrac2310^n+\dfrac13$ and $\dfrac2310^n+\dfrac43$?
A: Whenever I see something of the form $A + 2A\times\text{something of magnitude} + A\times \times\text{something of geometrically more magnitude}$, I figure it fits roughly into the form of $A(\text{something of mangitude} + 1)^2$.
So so $4_{2n}= \frac 49\times (10^{2n} -1)$ is roughly of magnitude $(10^n)^2$ and $8_{n} = 49\times (10^n -1)$ is roughly of magnitude $10^n$ then we should figure $4_{2n}+ 8_n + 4$ is fairly close so $\frac 49(10^n + 1)^2$.
And it turns it is VERY close (but slightly more) that it is too large to be a square of $\frac 23(10^n+1)$ but too small to be the next integer square.
(We might pause and consider that $\frac 23(10^n+1)$ need not be an integer.  That turns out not to be relevant.)
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To continue along your lines:
Express as $\sqrt{f(n)}=\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$
$=\frac 23\sqrt{(10^n)^2 + 2*10^n + 6}$
Notice that $n \ge  1$ then $(10^n + 1)^2 = 10^{2n} + 2*10^n  + 1 < (10^n)^2 + 2*10^n + 6 < 10^{2n} + 4*10^n + 4 = (10^n + 2)^2$
So  $(10^n)^2 + 2*10^n + 6$ is not a perfect square (but is an integer) so $\sqrt{2^{n+1} 5^n+100^n+6}$ is irrational and $\frac{2}{3} \sqrt{2^{n+1} 5^n+100^n+6}$ is not an integer.
