Write $x_n=22..244...45$ as sum of $2$ squares

Question

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.

Answer 1

You are on the right track. Write everything as a fraction with denominator 9 and use the fact that $10^n \equiv 1\pmod 3$ for all $n \geq 1$.

Answer 2

\begin{align*} x_{n} &= \left( \frac{10^{n}+a}{3} \right)^{2}+ \left( \frac{10^{n}+b}{3} \right)^{2} \\ &= \frac{2(10^{2n})+2(a+b)10^{n}+a^{2}+b^{2}}{9} \end{align*}

Solving $$\left \{ \begin{array}{ccc} a+b &=& 1 \\ a^{2}+b^{2} &=& 5 \end{array} \right.$$

$$(a,b)=(2,-1),(-1,2)$$

Answer 3

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