Let $a,b,c,m$ be integers, where $m>1$. Prove that if
$$a^n+bn+c\equiv0\pmod m$$for each natural number $n$, then $b^2\equiv0\pmod m$. Must $b\equiv0\pmod m$ also hold?
1. Given the congruence \( a^n + bn + c \equiv 0 \pmod{m} \) for each natural number \( n \), we start by considering specific values of \( n \).

2. For \( n = 1 \):
   \[
   a + b + c \equiv 0 \pmod{m}
   \]

3. For \( n = 2 \):
   \[
   a^2 + 2b + c \equiv 0 \pmod{m}
   \]

4. For \( n = 3 \):
   \[
   a^3 + 3b + c \equiv 0 \pmod{m}
   \]

5. We now have the following system of congruences:
   \[
   \begin{cases}
   a + b + c \equiv 0 \pmod{m} \\
   a^2 + 2b + c \equiv 0 \pmod{m} \\
   a^3 + 3b + c \equiv 0 \pmod{m}
   \end{cases}
   \]

6. Subtract the first congruence from the second:
   \[
   (a^2 + 2b + c) - (a + b + c) \equiv 0 \pmod{m}
   \]
   Simplifying, we get:
   \[
   a^2 + b - a \equiv 0 \pmod{m}
   \]
   \[
   b \equiv a - a^2 \pmod{m}
   \]

7. Subtract the second congruence from the third:
   \[
   (a^3 + 3b + c) - (a^2 + 2b + c) \equiv 0 \pmod{m}
   \]
   Simplifying, we get:
   \[
   a^3 + b - a^2 \equiv 0 \pmod{m}
   \]
   \[
   b \equiv a^2 - a^3 \pmod{m}
   \]

8. We now have two expressions for \( b \):
   \[
   b \equiv a - a^2 \pmod{m}
   \]
   \[
   b \equiv a^2 - a^3 \pmod{m}
   \]

9. Equate the two expressions for \( b \):
   \[
   a - a^2 \equiv a^2 - a^3 \pmod{m}
   \]
   Simplifying, we get:
   \[
   a - a^2 - a^2 + a^3 \equiv 0 \pmod{m}
   \]
   \[
   a - 2a^2 + a^3 \equiv 0 \pmod{m}
   \]
   \[
   a(a^2 - 2a + 1) \equiv 0 \pmod{m}
   \]
   \[
   a(a - 1)^2 \equiv 0 \pmod{m}
   \]

10. Since \( m > 1 \), the above congruence implies that either \( a \equiv 0 \pmod{m} \) or \( (a - 1)^2 \equiv 0 \pmod{m} \).

11. If \( a \equiv 0 \pmod{m} \):
    \[
    b \equiv 0 - 0 \equiv 0 \pmod{m}
    \]

12. If \( a \equiv 1 \pmod{m} \):
    \[
    b \equiv 1 - 1^2 \equiv 0 \pmod{m}
    \]

13. In both cases, we have \( b \equiv 0 \pmod{m} \). Therefore, \( b^2 \equiv 0 \pmod{m} \) is also true.

Conclusion:
\[
\boxed{b \equiv 0 \pmod{m}}
\]