Continuing with my current practice of finding mathematical problems to solve to keep my mind occupied, my most recent foray has been a BMO1993-r1 problem:
Here I begin by assigning $B=\sqrt{n}$ and $A$ to be the three most significant digits of $n$. Then playing around we get:
$ n=B^2=1001\times A+1 $
Which suggests we rearrange this (a common manoeuvre for such problems) as:
$ B^2 -1=1001\times A $
$ (B+1)(B-1)=1001\times A $
Which tells us we want to rearrange the RHS of this last equation as the product of two integers that differ by $2$. Also we can quickly eliminate the possibilities that the RHS is $1001\times 999$ and $1001\times 1002$ as one is not a square and the other has more than six digits.
That $B^2$ has six digits give constraints on possible values of $B$: $999\ge B \ge 317 $
The prime factorisation of $1001=7\times 11 \times 13$ and two of these factors must be factors of one of terms $(B+1)$ or $(B-1)$ and the remaining factor of the other term, so for some partition(s) $(p_1,p_2,p_3)$ of $(7,11,13)$ we may write:
$ (B+1)(B-1)=(p_1 \times p_2 \times U) \times (p_3 \times V) $
where the first barracked term on the RHS differs from the second by $\pm 2$.
So lets take $p_1=7$ and $p_2=11$, then we have:
$ (B+1)(B-1)=(77 \times U) \times (13 \times V) $
Then for the $B$s to be in range $U$ is one of $5,6,7,8,9,10,11,12$ and $77$ times these are $385,462 ,539,616,693,770,847,924$.
For a solution we need one or more of these to differ from a multiple of $13$ by $\pm 2$.
These are approximately $30,36,41,47,53,59,65,71$ $13$s or $390,468,533,611,689,767,845,923$.
We spot that the next to last of these differs from the next to last of the candidate for $U \times 77$ by $2$, and so provides a solution $B=846$, and $n=715716$.
There is no guarantee that this solution is unique, but then the question never asked for all, or a unique solution. Mechanising the heavy lifting give the fill list of solutions for $n$ as: $183184, 328329,528529,715716$.
