Write $x_n=22..244...45$ as sum of $2$ squares I've recently came across this problem and, although I've spent time looking for a solution, I don't have any interesting ideas.

Let the numbers 
  $$x_1=25$$
  $$x_2=2245$$
  $$x_3=222445$$
  and $x_n=22..244...45$ with $n$ digits $'2'$ and $n-1$ digits $'4'$
Prove that $x_n$ can be written  as a sum of two perfect squares for any natural number $n\ge1$.

The first impulse was to decompose the number:
$$ x_n = 2 \times \left(10^{2n-1}+\dots+10^n\right) + 4 \times \left(10^{n-1}+\dots+10\right)+ 5$$
$$ x_n = 2 \times \frac{10^{2n}-10}{9} + 2 \times \frac{10^{n}-10}{9} + 5$$
$$ x_n = \frac{20}{9} \left(10^{2n-1} - 1 + 10^{n-1} - 1\right) + 5$$
Anyway, this doesn't seem to lead to the good track. A piece of advice or a hint would be apreciated.
P.S. I guess that it's because of the Stack Exchange android app, but although I wrote something on the first line ("Hi there"), this doesn't appear in the post. This happened several times.

Since you specifically mentioned interesting ideas, I have posted this answer - see the last section
Outline
We exclude the case $25$ which is $3^2 + 4^2$.
Claim 
For all other numbers with $\color{blue}{n}$ $2$'s, $\color{blue}{n-1}$ $4$'s and $\color{blue}{one}$ $5$, they can be expressed as sum of $2$ squares, one of which is $\color{blue}{n-2}$ $6$'s.
Justification
As simplified by OP, $x_n = \frac{1}{9} \times \{20(10^{2n-1} + 10^{n-1}) + 5\}$
Let $t_n = 6 (1 + 10 + \ldots + 10^{n-2})$, then 
$x_n - t_n^2$ can be shown to be $(X + 3)^2/81$ where $X = 42 \times 10^{n-1}$.
The calculations got a little messy and so I am excluding them from here, but one can check, for $n = 2$ we will get $(420 + 3)/9 = 47$ and $47^2 + 6^2 = 2245$
Interesting
$$2245 = 47^2 + 6^2$$
$$222445 = 467^2 + 66^2$$
$$22224445 = 4667^2 + 666^2$$ and so on and that is exactly what we have shown sans the details of some calculations which can be filled in by anyone with basic algebra skills
