[ [decimal number system ]  $[$ equations in integers $]$

The number $x$ is such that $x^{2}$ ends in 001 (in the decimal number system).

Find the last three digits of the number $x$ (list all possible options).
$(x-1)(x+1)=x^{2}-1$ is divisible by $1000=2^{3} \cdot 5^{3}$. Therefore, both numbers $x-1$ and $x+1$ are even (then one of them is divisible by 4, and their product is divisible by 8), and one of them is divisible by 125, and thus by 250.

## Answer

$001,249,251,499,501,749,751,999$.