Written in a more formal way, proof that there are at the most $2$ numbers $n$ of six digits, that
$$n^2 \equiv n \mod 10^6$$
Research effort:
if $n^2 \equiv n \mod 10^6$ this means $10^6\mid n^2-n \rightarrow10^6\mid n(n-1)$
Given that $5^6\mid 10^6$ then $5^6\mid n(n-1)$ whitch means $5^6\mid n$ or $5^6\mid n-1$
Same for $2^6$, Given that $2^6\mid 10^6$ then $2^6\mid n(n-1)$ whitch means $2^6\mid n$ or $2^6\mid n-1$
By the chinese remainder theorem I got assured $4$ solutions, but I tried to solve the systems and it's not a very happy thing to do, so I think I did something wrong somewhere.
What do you think?
If $\displaystyle2^6|(n-a)$ and $5^6|(n-a),$
$\displaystyle$ lcm$(2^6,5^6)|(n-a)$
As $(2,5)=1,(2^6,5^6)=(2,5)^6=1$
$\implies$ lcm$(2^6,5^6)=(2^6\cdot5^6)=10^6\iff n\equiv a\pmod{10^6}$
Here $a=0,1$