This answer explores further the problem described this one, where base-10 numbers that become squares when the digits are repeated twofold are identified (e.g. $13223140496\to1322314049613223140496=36363636364^2$).
In any such square concatenation, the number of digits in each "half" of the square is a value of $n$ for which $10^n+1$ has a squared prime factor. For instance, $10^{11}+1$ is divisible by $11^2$, so there are eleven-digit numbers that become square when concatenated twofold, like the one noted above.
We can identify various solutions for $n$ corresponding to a given squared prime factor by the following process.
Select a value of $m$ such that $p|10^m+1$. Such a value of $m$ will be an odd number times half the base-ten repeating decimal period length of $1/p$; the period length itself must be even for the selected $p$ or no solution exists.
Render $10^m=K-1$ where $K=10^m+1$ and raise to the power of $p$, writing the right side of $(K-1)^p$ as a polynomial in $K$ via the Binomial Theorem.
Each term in that binomial- power expansion is constructed to be a multiple of $p^2$ except the $-1$ term at the end, so $10^{mp}+1$ is a certified multiple of $p^2$.
Results for small primes $p$ suggest that if we choose the minimal value of $m$ for a given factor $p$, then $n=mp$ derived in this way is the minimal solution to $p^2|10^n+1$. For instance, $p=11$ pairs with $m=1$ (because $10^1+1$ is $11$ itself), giving $10^{11}+1$ as the smallest augmentedpower of ten divisible by $11^2$. Similarly, $10^8+1$ is the smallest augmented power of ten divisible by $17$, and the solution found by the procedure above, $10^{136}+1$, is the minimal augmented power of ten divisible by $17^2$.
All is as expected until we hit $\color{blue}{p=487}$. The minimal $m$ value is $243$, so we we find from the above lifting procedure that that $487^2|10^{118431}+1$ ... but OEIS sequence A086981 reports that instead $487^2|10^{243}+1$ as the minimal solution for that factor -- that is, $10^m+1$ instead of $10^{mp}+1$. The minimal solution for that squared prime factor slips in before using the binomial power expansion, not after as is usually the case.
Such a counterexample is connected with the fact that any residue $r\bmod p$, maps into $p$ residues $\bmod p^2$. One of these mapped residues is $\equiv r^p$, and this one is a $p$-power residue $\bmod p^2$. The other residues $\bmod p^2$ are not $p$-power residues.
For instance, with $p=7$ we render
$10^7\equiv3^7=2187\equiv31\bmod 49,$
so $31$ is a seventh-power residue $\bmod 49$ and other residues congruent to it $\bmod 7$, including $10$, are not seventh-power residues.
With other prime factors that allow a solution to $p|10^m+1$ (and thus also $p^2|10^n+1$), we get the following from residue $10\bmod p$:
$p=11\to10^{11}\equiv120\bmod11^2$
$p=13\to10^{13}\equiv23\bmod13^2$
$p=17\to10^{17}\equiv214\bmod17^2$
...
$p=463\to10^{463}\equiv12974\bmod463^2$
$\color{blue}{p=487\to10^{487}\equiv10\bmod487^2}$
So unlike all smaller primes in scope, $p=487$ renders $10$ a $p$-power residue $\bmod p^2$, causing an extra factor of $p$ to appear in $10^m+1$ for appropriates value of $m$. Hence for this particular case, we did not have to multiply the initial exponent $m=243$ by $487$ to get an augmented power of ten divisible by $487^2$. If we do choose to multiply the exponent, we find that $10^{118341}+1$ is divisible by $487^3$.
The documentation of OEIS A086981 reports no other counterexamples below $1000$. Larger counterexamples are expected to become rare because the $p$-power residue condition becomes less probable for larger primes $p$.
