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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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    $\begingroup$ what is the source of the problem $\endgroup$ – Albus Dumbledore Aug 30 '20 at 14:22
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    $\begingroup$ Are you searching for only positive primes or also negative? $\endgroup$ – Alessandro Cigna Aug 30 '20 at 14:54
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    $\begingroup$ @aryanbansal $|x|=3$ does not imply $x=3$. $\endgroup$ – TheSilverDoe Aug 30 '20 at 15:16
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    $\begingroup$ I did all the possible cases with lot of counts, and the answer is no, but I’m searching for a faster way $\endgroup$ – Alessandro Cigna Aug 30 '20 at 15:26
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    $\begingroup$ See artofproblemsolving.com/community/c6h392562p3159787 $\endgroup$ – Sil Aug 30 '20 at 20:00
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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 $p_1<p_2<p_3$) 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.
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  • $\begingroup$ Looks great! Can you elaborate on why If $|p_4-p_1|= 6$ then we have $|p_4-p_2|=1$...? $\endgroup$ – mathcounterexamples.net Aug 30 '20 at 16:59
  • $\begingroup$ $Q(x) =a(x-p_1)(x-p_2)(x-q)+3$ so $-3=Q(p_4)= a(p_4-p_1)(p_4-p_2)(p_4-q)+3$. @mathcounterexamples.net $\endgroup$ – Aqua Aug 30 '20 at 17:18
  • $\begingroup$ @Aqua Indeed thanks! $\endgroup$ – mathcounterexamples.net Aug 30 '20 at 17:22
  • $\begingroup$ @Aqua can you please elaborate further? Because q could be not an integer as far as i'm concerned. $\endgroup$ – Foorgy Infifcio Aug 31 '20 at 19:33

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