Linear Function For Primes Let $f:Z \to F_p$ be a function with p>3, p being a prime. F has the following property:
$$f(\alpha+\beta)+f(\alpha+2\beta)+f(\alpha+3\beta)=0 $$for$$ \alpha∈Z, \beta∈Z+. $$
Prove f(n) = 0 for all $n \in Z$.
Is $F_p$ simply the field of prime numbers or the quotient ring? since field of the quotient ring have to be bigger or equal to 0, then the result is trivial for f(n)=0? am i missing something?

(I assume that $F_p$ is the field with $p$ elements, aka $\mathbb Z/p\mathbb Z$)
Fix $\beta=1$:
$\alpha=0 \implies f(1)+f(2)+f(3)=0$
$\alpha=1 \implies f(2)+f(3)+f(4)=0 \implies f(1)=f(4)$
$\alpha=2 \implies f(3)+f(4)+f(5)=0 \implies f(2)=f(5)$
$\alpha=3 \implies f(4)+f(5)+f(6)=0 \implies f(3)=f(6)$
In general, $f(n)=f(n \bmod 3)$.
Fix $\beta=3$:
$\alpha=0 \implies f(3)+f(6)+f(9)=0 \ \ \implies 3f(3)=0 \implies f(3)=0$
$\alpha=1 \implies f(4)+f(7)+f(10)=0 \implies 3f(4)=0 \implies f(4)=0$
$\alpha=2 \implies f(5)+f(8)+f(11)=0 \implies 3f(5)=0 \implies f(5)=0$
Thus, $f(n)=f(n \bmod 3)=0$.
