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Let $f:F\to F$ be defined by $f(x)=x^3$ where $F=\mathbb{Z}/p\mathbb{Z}$ for some prime $p$. Prove that $f$ is linear if and only if $p=3$.

My attempt:

First suppose $p=3$. Then $$f(x+y)=x^3+3x^2y+3xy^2+y^3=x^3+y^3=f(x)+f(y)$$ and $$f(\lambda x) = \lambda^3 x^3 = \lambda x^3 = \lambda f(x)$$

Therefore $f$ is linear.

Now suppose $p \neq 3$. I'm not sure what to do from here, how do I do this?

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    For one thing, you also need to assume that $p\neq 2$ since the map is also linear then. Once you do that, consider $f(2)$. – 2017-01-02

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If $a^3=f(a \cdot 1)=af(1)=a$ for every $a \in \Bbb F_p$, then it holds in particular for a generator $g$ of $\Bbb F_p^*$, which is a cyclic group. Think about the order of $g$, noticing that $g^2=1$. Can you take it from here?

Other idea: if $a=2$, you get $a^3 = 8 \equiv 2 = a \pmod p$. Can you take it from here?