True or false? Squares

The concatenation operation is equivalent to multipling $n$ by $C=10^d+1$ , where $d$ is the number of digits of $n$ . Note that $C>n$ (it has one more digit).

We need $Cn=x^2$ for some integer $x$ .

If $C$ has no square factors, then all of the prime factors of $C$ have to be prime factors of $n$ . If this is the case, then $n$ is at least as big as $C$ ; but we've already established that $C>n$ , so this is impossible.

For a concrete example, let's try to find a $3$ -digit $n$ that works. We have $C=1001=7\cdot11\cdot13$ , so

$7\cdot11\cdot13\cdot n = x^2$

Each of $7,11,13$ must be factors of $n$ ; but the smallest positive number with these factors is $1001$ ; contradiction.

This means we're looking for a $C$ with a square factor. The first one we find is $C=10^{11}+1=11^2\cdot23\cdot4093\cdot8779$ , so $n$ has to have $11$ digits.

$11^2\cdot23\cdot4093\cdot8779\cdot n=x^2$

We could take $n=23\cdot4093\cdot8779=826446281$ , which gives a square, but this only has $9$ digits. To get around this, we can multiply by a small square number to increase the number of digits.

$n=16\cdot23\cdot4093\cdot8779=\boxed{13223140496}$ is the smallest possible $n$ with $11$ digits.

In full, $\overline{nn}=Cn=11^2\cdot23\cdot4093\cdot8779\cdot16\cdot23\cdot4093\cdot8779 = (4\cdot11\cdot23\cdot4093\cdot8779)^2=36363636364^2$

Two quick notes:

• the nice form of $x$ came as a surprise - I wonder if it means there's a different route to this solution?
• these are apparently called "biperiod squares"

See https://oeis.org/A102567 for more information.