Show that $2n+2$ - digit number: $11...122...25$ (there are $n$ of $11...1$ and $(n+1)$ of $22...2$) is square of natural number.
I tried to solve it by induction
First for $n=0$ we have $25=5^2$
We assume that number $11...122...25 = k^2$ (there are $n$ of $11...1$ and $(n+1)$ of $22...2$)
And need to show that number $111...1222...25$ (there are $(n+1)$ of $11...1$ and $(n+2)$ of $22...2)$ is also a square of natural, but here I have prolem to show it.
Let $S = 11...122...25$ (the number you defined)
Let $A = 11...1$ (n of $1$s)
$S = A\cdot 10^{n+2} + 2\cdot A\cdot 10^2 + 25$
Substitute $A = \frac{10^n-1}{9}$ into the equation:
$$S = \frac{10^n-1}{9}\cdot 10^{n+2} + 2\cdot \frac{10^n-1}{9}\cdot 10^2 + 25$$
$$\Leftrightarrow S = \frac{(10^n-1)\cdot 10^{n+2}}{9} + \frac{2\cdot(10^n-1)\cdot 10^2 + 9\cdot25}{9}$$
$$\Leftrightarrow S = \frac{(10^n-1)\cdot 10^{n+2}+2\cdot(10^n-1)\cdot 10^2 + 225}{9}$$
$$\Leftrightarrow S = \frac{10^{2n+2}-10^{n+2}+2\cdot10^{n+2}-2\cdot 10^2 + 225}{9}$$
$$\Leftrightarrow S = \frac{10^{2n+2}+10^{n+2}+25}{9}$$
$$\Leftrightarrow S = \frac{(10^{n+1}+5)^2}{9}$$
$$\Leftrightarrow S = (\frac{10^{n+1}+5}{3})^2$$
$ 10^{n+1} + 5 \equiv 0 \pmod3 $
Hence S is square of a natural number.
Hint $\ $ Below is a sketch of an inductive proof.
If $\,\ f_k = 11\ldots1122\ldots2225\, =\, 33\ldots335^2 = n^2\ \,$ then $\begin{eqnarray} 11\ldots 1122\ldots222500 \,&=&\,\ \ \, 100\,n^2\phantom{\large I^I}\\ \quad\quad\! {-}10\ldots 000500 \,&=&\, -300\, n\\ +225 \,&=&\ {+}225\\ \hline \!\!\!f_{k+1}=\,\ 11\ldots11\color{#c00}12\ldots222\color{#c00}{22}5 \,&=&\ (10\,n\!-\!15)^2 =\, 33\ldots33\color{#c00}35^2 \end{eqnarray}$
