There are only two six-digit integers $N$, each greater than $100,000$. for which $N^2$ has $N$ as its final six digits There are only two six-digit integers $N$, each greater than 100,000 for which $N^2$ has $N$ as its final six digits (or $N^2-N$ is divisible by $10^6$). What are these two numbers?
Is the problem solvable by the Chinese Remainder Theorem? If so, how?

Yes, we can solve $N^2-N\equiv0\bmod10^6$ with the Chinese remainder theorem
to get $N\equiv0 $ or $1\bmod 2^6=64$ and $N\equiv0$ or $ 1\bmod5^6=15625$.
We don't want the solutions $N\equiv0\bmod2^6$ and $5^6$ or $N\equiv1\bmod2^6 $ and $5^6$.
We want the solutions $N\equiv1\bmod 2^6$ and $N\equiv0\bmod5^6$ and vice versa.
The extended Euclidean algorithm gives $15625=244\times64+9$ and $64=7\times9+1$,
so $1=1709\times64-7\times15625$.
Thus, one solution is $1709\times64$, and the other is $(64-7)\times15625$.
