125. For which integer values of $x$ does the expression $\sqrt{x^{2}-x+1}$ represent an integer?
Solution. The given expression is an integer if $x^{2}-x+1=n^{2}$ (1), where $n-$ is a natural number. Multiplying equation (1) by 4, we get:

$$
4 x^{2}-4 x+4=4 n^{2}, \text { i.e. }(2 x-1)^{2}=(2 n)^{2}-3 .
$$

We have obtained that the squares of natural numbers

$$
4 x^{2}-4 x+1=(2 x-1)^{2} \text { and } 4 x^{2}-4 x+4=(2 n)^{2}
$$

differ by 3. Therefore, the first one equals 1. Hence, $x=0$ or $x=1$.

Answer: $x=0 ; x=1$.