Problem 1255

1. First, we note that $840 = 3 \cdot 5 \cdot 7 \cdot 8$. We need to find the integers $n$ such that $0 \leq n < 840$ and $840$ divides $n^8 - n^4 + n - 1$. This means we need $n^8 - n^4 + n - 1 \equiv 0 \pmod{840}$.

2. We will consider the congruences modulo $3$, $5$, $7$, and $8$ separately.

3. Modulo 3:
- Since $n^2 \equiv 1 \pmod{3}$ for $n$ such that $\gcd(n, 3) = 1$, we have:
$$n^8 \equiv 1 \pmod{3}, \quad n^4 \equiv 1 \pmod{3}$$
$$n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{3}$$
Therefore, $n \equiv 1 \pmod{3}$.

4. Modulo 5:
- Since $n^4 \equiv 1 \pmod{5}$ for $n$ such that $\gcd(n, 5) = 1$, we have:
$$n^8 \equiv 1 \pmod{5}, \quad n^4 \equiv 1 \pmod{5}$$
$$n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{5}$$
Therefore, $n \equiv 1 \pmod{5}$.

5. Modulo 8:
- Since $\varphi(8) = 4$, we have $n^4 \equiv 1 \pmod{8}$ for $n$ such that $\gcd(n, 8) = 1$, we have:
$$n^8 \equiv 1 \pmod{8}, \quad n^4 \equiv 1 \pmod{8}$$
$$n^8 - n^4 + n - 1 \equiv 1 - 1 + n - 1 \equiv n - 1 \pmod{8}$$
Therefore, $n \equiv 1 \pmod{8}$.

6. Modulo 7:
- Since $n^6 \equiv 1 \pmod{7}$, we have:
$$n^8 \equiv n^2 \pmod{7}, \quad n^4 \equiv n^2 \pmod{7}$$
$$n^8 - n^4 + n - 1 \equiv n^2 - n^2 + n - 1 \equiv n - 1 \pmod{7}$$
Therefore, $n \equiv 1 \pmod{7}$ or $n \equiv 3 \pmod{7}$.

7. Combining the congruences using the Chinese Remainder Theorem:
- For $n \equiv 1 \pmod{3}$, $n \equiv 1 \pmod{5}$, $n \equiv 1 \pmod{8}$, and $n \equiv 1 \pmod{7}$:
$$n \equiv 1 \pmod{840}$$
This gives us $n = 1$.

- For $n \equiv 1 \pmod{3}$, $n \equiv 1 \pmod{5}$, $n \equiv 1 \pmod{8}$, and $n \equiv 3 \pmod{7}$:
Using the Chinese Remainder Theorem, we find another solution $n$ in the interval $0 \leq n < 840$.

8. Therefore, there are two solutions in total.

The final answer is $\boxed{2}$.