$[\underline{\text { Modular Arithmetic (other) }}]$

For which $n$ is the number $n^{2001}-n^{4}$ divisible by 11?
According to Fermat's Little Theorem, \( n^{2001} - n^{4} \equiv n - n^{4} = n(1 - n)(n^{2} + n + 1) \pmod{11} \). \( n^{2} + n + 1 \equiv n^{2} - 10n + 25 - 2 = (n - 5)^{2} - 2 \pmod{11} \). Checking the remainders shows that a square cannot give a remainder of 2 when divided by 11.

## Answer

For \( n \equiv 0,1 \pmod{11} \).