WakeUp wrote: Three students write on the blackboard next to each other three two-digit squares. In the end, they observe that the 6-digit number thus obtained is also a square. Find this number! Let $a^2,b^2,c^2$ be the three squares (with $a,b,c\in\{4,5,6,7,8,9\}$) and $n^2=\overline{a^2b^2c^2}$. So $\sqrt{10000(a^2+1)}>n\ge 100a+4$ (the last $4$ is in order to have a two-digits square in the rightmost position of $n^2$). If $a=4$, we get $n\in[404,412]$ and just eight tests to get $408^2=166464$ (I did not count $410^2=168100$, since then the rightmost square does not have two digits). If $a=5$, we get $n\in[504,509]$ and just six tests and no solution; If $a=6$, we get $n\in[604,608]$ and just five tests and no solution; If $a=7$, we get $n\in[704,707]$ and just four tests and no solution; If $a=8$, we get $n\in[804,806]$ and just three tests to get $804^2=646416$; If $a=9$, we get $n\in[904,905]$ and just two tests and no solution. And so, two solutions: $\boxed{n\in\{166464,646416\}}$.

Understood, thanks @below.

huashiliao2020 wrote: I don't understand why you did +4.. shouldn't it be +10b+c? No. $n^2=\overline{a^2b^2c^2}$ does not mean $n=\overline{abc}$ But if $n=\overline{a0k}$ with $k<4$ then two rightmost digits of $n^2$ are $00,01,04,09$ and so not a two-digits square.