Is there a cubic $Q(x)\in \mathbb{Z}[x]$ so that $|Q(p_1)|=|Q(p_2)|=|Q(p_3)|=|Q(p_4)|=3$, where $p_1, p_2, p_3, p_4$ are distinct primes? https://ift.tt/eA8V8J

Is there a cubic $Q(x)\in \mathbb{Z}[x]$ so that $|Q(p_1)|=|Q(p_2)|=|Q(p_3)|=|Q(p_4)|=3$, where $p_1, p_2, p_3, p_4$ are distinct primes?

Clearly there must be at least one $Q(p_i)=3$ and at least one $Q(p_j)=-3$ (otherwise there will be 4 roots of a third degree polynomial)

Lets suppose that $Q(p_1) = 3$ and $Q(p_2) = -3$.

$Q(p_1) - Q(p_2)/ (p_1-p_2) = n$ where $n \in \mathbb{Z}$

The dividers of $6$ are $1, 2, 3, 6$. $(p_1-p_2) \in \{1, 2, 3, 6\}$

That’s what I’ve got so far.

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Foorgy Infifcio