To determine the function \( f(n) \) where \( f(n) \) is the remainder when \( n^8 + n^{16} \) is divided by \( 5 \), we first investigate this function for all possible values of \( n \) modulo \( 5 \). Specifically, we consider \( n \equiv 0, 1, 2, 3, \) and \( 4 \pmod{5} \).
### Case Analysis
**Case 1: \( n \equiv 0 \pmod{5} \)**
\[
n^8 \equiv 0^8 \equiv 0 \pmod{5} \quad \text{and} \quad n^{16} \equiv 0^{16} \equiv 0 \pmod{5}
\]
Thus, \( f(0) \equiv 0 + 0 \equiv 0 \pmod{5} \).
**Case 2: \( n \equiv 1 \pmod{5} \)**
\[
n^8 \equiv 1^8 \equiv 1 \pmod{5} \quad \text{and} \quad n^{16} \equiv 1^{16} \equiv 1 \pmod{5}
\]
Thus, \( f(1) \equiv 1 + 1 \equiv 2 \pmod{5} \).
**Case 3: \( n \equiv 2 \pmod{5} \)**
Using Fermat's Little Theorem, \( 2^4 \equiv 1 \pmod{5} \):
\[
2^8 = (2^4)^2 \equiv 1^2 \equiv 1 \pmod{5} \quad \text{and} \quad 2^{16} = (2^8)^2 \equiv 1^2 \equiv 1 \pmod{5}
\]
Thus, \( f(2) \equiv 1 + 1 \equiv 2 \pmod{5} \).
**Case 4: \( n \equiv 3 \pmod{5} \)**
Using Fermat’s Little theorem, \( 3^4 \equiv 1 \pmod{5} \):
\[
3^8 = (3^4)^2 \equiv 1^2 \equiv 1 \pmod{5} \quad \text{and} \quad 3^{16} = (3^8)^2 \equiv 1^2 \equiv 1 \pmod{5}
\]
Thus, \( f(3) \equiv 1 + 1 \equiv 2 \pmod{5} \).
**Case 5: \( n \equiv 4 \pmod{5} \)**
Using Fermat's Little Theorem, \( 4^4 \equiv 1 \pmod{5} \):
\[
4^8 = (4^4)^2 \equiv 1^2 \equiv 1 \pmod{5} \quad \text{and} \quad 4^{16} = (4^8)^2 \equiv 1^2 \equiv 1 \pmod{5}
\]
Thus, \( f(4) \equiv 1 + 1 \equiv 2 \pmod{5} \).
### Calculation of \( f(0) + f(1) + f(2) + f(3) + f(4) \)
Adding the individual results together:
\[
f(0) + f(1) + f(2) + f(3) + f(4) = 0 + 2 + 2 + 2 + 2 = 8
\]
Thus, the final answer is:
\[
\boxed{8}
\]
For an integer $n,$ let $f(n)$ be the remainder when $n^8 + n^{16}$ is divided by $5.$ Compute $f(0) + f(1) + f(2) + f(3) + f(4).$
1 answer