A mythical question about the fine-structure constant inverse approximation, $137$, and the telephone number for information, $411$.

Question

For which positive integers $a$, and $b$, does $a^2 + b^2$ divide the concatenation of $a$ and $b$ that is $ab$?

Eg, $4^2 + 11^2 = 137$, which divides $411 = 3 X 137$.

Is a there a mathematical name for the condition on integers $a$, $b$?

Answer 1

Let's write $a\|b$ for the concatenation of $a$ and $b$, e.g., $142\|857=142857$. Then the special case $a\|b=a^2+b^2$ is tabulated here. The first few terms are

$101, 1233, 8833, 10001, 10100, 990100, 1000001, 5882353, 94122353, 99009901, 100000001$, $100010000, 1765038125, 2584043776, 7416043776, 8235038125, 9901009901, 10000000001$, $48600220401, 116788321168, 123288328768, 601300773101, 876712328768, 883212321168$, $990100990100, 999900010000, 1000000000001, 1000001000000$.

The accompanying text begins,

The condition $x^2 + y^2 = 10^kx + y$ is equivalent to $(2x-10^k)^2 + (2y-1)^2 = 10^{2k} + 1$, so to find these sequence elements it is necessary to write $10^{2k} + 1$ as the sum of two squares.

There is a link to A. van der Poorten, K. Thomsen, and M. Wiebe, A curious cubic identity and self-similar sums of squares, The Mathematical Intelligencer, v.29(2), pp. 69-73, June 2007.

EDIT: Now, let's look at the general case, $a\|b=(a^2+b^2)r$. We write this as $$10^ka+b=(a^2+b^2)r$$ and after a little algebra we get $$(2ar-10^k)^2+(2br-1)^2=10^{2k}+1$$ so again it's a question of writing $10^{2k}+1$ as a sum of two squares. Let's do a couple of examples.

$\mathbf{k=2}$, $(2ar-100)^2+(2br-1)^2=10001=100^2+1^2=76^2+65^2$. Taking first $100^2+1^2$, we get $$2br-1=1,\quad2ar-100=100$$ so $br=1$, $ar=100$. The only solution is $r=1$, $b=1$, $a=100$, and $100\|01=100^2+01^2$.

Now using $76^2+65^2$, we get $$2br-1=65,\quad2ar-100=\pm76$$ from which we get $$br=33,\quad ar=88{\rm\ or\ }12$$ We get solutions

$r=1$, $b=33$, $a=88$: $88\|33=88^2+33^2$
$r=1$, $b=33$, $a=12$: $12\|33=12^2+33^2$
$r=3$, $b=11$, $a=4$: $4\|11=3(4^2+11^2)$ (which started this whole question)
$r=11$, $b=3$, $a=8$: $803=11(8^2+03^2)$.

Let's do $\mathbf{k=4}$. We get $10^8+1=10000^2+1^2=8824^2+4705^2$.

$2br-1=1$, $2ar-10000=10000$ leads to $r=1$, $b=1$, $a=10000$ and the trivial $10000\|0001=10000^2+0001^2$.

$2br-1=4705$, $2ar-10000=\pm8824$ leads to
$br=2353=13\times181$, $ar=9412=4\times13\times181$, or $ar=588=3\times4\times49$.

We get $r=1$, $b=2353$, $a=9412$: $9412\|2353=9412^2+2353^2$
$r=1$, $b=2353$, $a=588$: $588\|2353=588^2+2353^2$
$r=13$, $b=181$, $a=724$: $7240181=13(724^2+181^2)$
$r=181$, $b=13$, $a=52$: $520013=181(52^2+13^2)$
$r=2353$, $b=1$, $a=4$: $40001=2353(4^2+1^2)$.