# 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?

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.

• what is the source of the problem – Albus Dumbledore Aug 30 at 14:22
• Are you searching for only positive primes or also negative? – Alessandro Cigna Aug 30 at 14:54
• @aryanbansal $|x|=3$ does not imply $x=3$. – TheSilverDoe Aug 30 at 15:16
• I did all the possible cases with lot of counts, and the answer is no, but I’m searching for a faster way – Alessandro Cigna Aug 30 at 15:26
• – Sil Aug 30 at 20:00

Clearly values $$Q(p_i)$$ can not be all the same since third degree polynomial can take only 3 times the same value. Then we have a following cases:

• Suppose $$Q(p_1)= Q(p_2)= Q(p_3)=3$$ and $$Q(p_4)=-3$$, so $$Q(x) = a(x-p_1)(x-p_2)(x-p_3)+3$$ and thus $$-6 = a(p_4-p_1)(p_4-p_2)(p_4-p_3)$$ Since primes are all different (say $$p_1) we have: $$6 = |a||(p_4-p_1)||(p_4-p_2)||(p_4-p_3)|\geq 1\cdot 1\cdot 2\cdot 3 = 6$$ and this means that $$|p_4-p_1|$$ and $$|p_4-p_3|$$ are odd so $$p_4=2$$ which is impossible or $$p_1=p_3 = 2$$ which is again impossible.

• If $$Q(p_1)= Q(p_2)= Q(p_3)=-3$$ and $$Q(p_4)=3$$ we proceed similarly as in first case.

• Suppose $$Q(p_1)= Q(p_2)=3$$ and $$Q(p_3)=Q(p_4)=-3$$, then we have: $$p_4-p_1\mid Q(p_4)-Q(p_1) =-6$$ and similarly for all other pairs, so

$$|p_4-p_1|,|p_4-p_2|,|p_3-p_1|,|p_3-p_2|\in\{1,2,3,6\}$$

• If $$|p_4-p_1|= 6$$ then we have $$|p_4-p_2|=1$$ so $$p_2=2$$ and $$p_4 =3$$ and $$p_1=9$$ or $$p_2=3$$ and $$p_4 =2$$ and $$p_1=8$$. A contradiction. Similarly we see that all absolute differences can not be 6. So two differences must be the same.

If two of them are 3 or 1 then we have two primes to be 2. Impossible.

If two of them are 2 then we have two subcases:

• $$|p_4-p_1|= |p_4-p_2|= 2$$ then $$|p_2-p_1|=4$$ but $$4\nmid 6$$.
• $$|p_4-p_1|= |p_3-p_2|= 2$$ then $$|p_3-p_1|$$ and $$|p_4-p_2|$$ are odd so we have again two primes equal 2. A contradiction again.
• Looks great! Can you elaborate on why If $|p_4-p_1|= 6$ then we have $|p_4-p_2|=1$...? – mathcounterexamples.net Aug 30 at 16:59
• $Q(x) =a(x-p_1)(x-p_2)(x-q)+3$ so $-3=Q(p_4)= a(p_4-p_1)(p_4-p_2)(p_4-q)+3$. @mathcounterexamples.net – Aqua Aug 30 at 17:18
• @Aqua Indeed thanks! – mathcounterexamples.net Aug 30 at 17:22
• @Aqua can you please elaborate further? Because q could be not an integer as far as i'm concerned. – Foorgy Infifcio Aug 31 at 19:33