<s>[INST] How to prove $(6666\ldots66)^2 + 8888\ldots88 = 4444\ldots44$ (with $n$ 6s and 8s, $2n$ 4s) How to prove that, if $n$ is a positive integer, then
$$
  (\underbrace{666 \ldots 6}_{n \text{ copies of } 6})^2 + 
  \underbrace{888 \ldots 8}_{n \text{ copies of } 8} = 
  \underbrace{444 \ldots 4}_{2n \text{ copies of } 4}?
$$
 [/INST] Given $$\underbrace{(666666666666\ldots)^2}_{n~\text{times}}+\underbrace{(888888888888\ldots)}_{n~\text{times}}$$
Now we can write $$\underbrace{666666666666\ldots}_{n~\text{times}} = (6+6\cdot 10+6\cdot 10^2+\cdots+6\cdot 10^{n-1})$$
$$\displaystyle =6\left[\frac{10^n-1}{10-1}\right] = \frac{2}{3}\left[10^n-1\right]$$
Similarly  we can write $$\underbrace{88888888888\ldots}_{n~\text{times}} = (8+8\cdot 10+8\cdot 10^2+\cdots+8\cdot 10^{n-1})$$
$$\displaystyle =8\left[\frac{10^n-1}{10-1}\right] = \frac{8}{9}\left[10^n-1\right]$$
So we get $$\displaystyle \left\{\frac{2}{3}\left[10^n-1\right]\right\}^2+\frac{8}{9}\left[10^n-1\right] = \frac{4}{9}(10^n-1)^2+\frac{8}{9}(10^n-1)$$
So we get $$\displaystyle = \frac{4}{9}(10^n-1)\cdot \left[10^n-1+2\right] = \frac{4}{9}(10^{2n}-1)$$
So  we can write  $$\displaystyle \frac{4}{9}(10^{2n}-1) = \underbrace{(444444444444\ldots)}_{2n~\text{times}}$$
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