We looked at three puzzles, and chose Problem 346. After some experiments, we conjectured that the only square in the list is 9, and tried expressing numbers like $88889$ in several different ways, starting from $999=10^4-1$.
We looked at three puzzles, and chose Problem 346. After some experiments, we conjectured that the only square in the list is 9, and tried expressing numbers like $88889$ in several different ways, starting from $999=10^4-1$.
Since $9\ldots 9=10^n-1$, write $8\ldots 89$ as $1+8(10^n-1)/9$. If $1+8(10^n-1)/9$ is square then $9+8(10^n-1)=m^2$,
so $m^2-9=(m-3)(m+3)$ gives an impossible factorization of $8(10^n-1)$ (so just prove that).