Conjecture : Let $k$ be a non-negative integer. Then , there is a positive integer $j$ , such that $10^j+k$ is NOT squarefree.
The conjecture is true upto $k=10^6$. We can always find a positive integer $j\le 421$ as desired.
If we demand that $10^j+k$ is not cubefree, then still up to $k=17493$ L, there is a $j\le 6376$ as desired.
I am however currently mainly interested in the non-squarefree case.
Another approach (which I have not yet tried) is that we determine the $k$ for which $4\mid 10^j+k$ , $9\mid 10^j+k$ , $25\mid 10^j+k$ is possible and hope that we can "cover" all positive integers.
If someone solves this problem for which we need discrete logarithms, an answer would be appreciated and I would gladly upvote and accept it. The non-cubefree case would be a nice bonus.
Best Answer
Artin's conjecture on primitive roots asserts that there are infinitely many primes $p$ for which $10$ is a primitive root, that is, for which $10,10^2,\dots,10^{p-1}$ cover all nonzero residue classes modulo $p$. Artin's conjecture is known to follow from a suitable version of the generalized Riemann hypothesis.
While it is not usually stated this way, the heuristic that leads to Artin's conjecture also leads to the assertion that there are infinitely many primes $p$ for which $10$ is a primitive root modulo $p^2$, and the proof assuming GRH can be adapted to justify this assertion. (If $10$ is a primitive root modulo $p$, the only way it could fail to be a primitive root modulo $p^2$ is if $10^{p-1}\equiv 1\pmod{p^2}$, that is, if $p$ were a base-$10$ Wieferich prime; such primes are believed to be far rarer than the primes for which $10$ is a primitive root.)
With all this in mind: choose $p$ such that $10$ is a primitive root modulo $p^2$. Then there exists $1\le j\le p(p-1)$ such that $10^j \equiv -k\pmod{p^2}$. In other words, $p^2$ divides $10^j+k$, which is therefore not squarefree. (Indeed, even a single such prime $p$ would produce infinitely many nonsquarefree numbers $10^{j+mp(p-1)}+k$ for $m=0,1,2,\dots$.)
If $10$ is a primitive root modulo $p^2$, then it is automatically a primitive root modulo $p^r$ for all $r\ge3$ as well. Therefore the same argument shows that there will be exponents $j$ such that $10^j+k$ is not cubefree, not $4$th-power-free, etc.
General remark: mathematicians don't know any nice arithmetic formula that always produces squarefree numbers (just like we don't know any nice arithmetic formula that always produces primes). So $10^j+k$ always being squarefree for some fixed $k$ would be incredibly unexpected.