Let $p=x^4 + 25x^2 + 125$ be a prime. Prove that $2$ is a quintic residue $\pmod p$, and therefore $y^5=2\pmod p$ is solvable.

A similar example was first conjectured by Euler:

If $p=x^2 + 27$ is a prime, then $2$ is a cubic residue $\pmod p$, $y^3=2\pmod p$ is solvable.


Digging into this more, my conjecture holds for $x < 1000$ plus more values, and has many common properties with Euler's cubic reciprocity conjecture involving the polynomial $x^2 + 27$.

According to this page on cubic reciprocity, $2$ is a cubic residue $\pmod p$ if and only if $p=x^2 + 27y^2$.

Similarly, if $p=x^4 + 25x^2y^2 + 125y^4$, then $2$ is a quintic residue $\pmod p$. This also seems to hold.

Besides $2$ being a cubic and quintic residue of $x^2 + 27$ and $x^4 + 25x^2 + 125$, respectively, $x$ also holds the same property.

If $p=x^2 + 27$, then $x$ is a cubic residue $\pmod p$.

If $p=x^4 + 25x^2 + 125$, then $x$ is a quintic residue $\pmod p$.

Euler's logic and reasoning about his conjectures on cubic residues could likely be applied to quintic residues.

  • 1
    $\begingroup$ This looks hard. Has Euler's conjecture been resolved? $\endgroup$ – Robert Lewis Aug 12 '18 at 16:41
  • 1
    $\begingroup$ @RobertLewis Yes, by Gauss if I recall correctly. $\endgroup$ – Lord Shark the Unknown Aug 12 '18 at 17:17
  • $\begingroup$ @LordSharktheUnknown: I suspected that might be the case. Thanks! $\endgroup$ – Robert Lewis Aug 12 '18 at 17:19

This is a special case of a result due to Emma Lehmer (see this article). In (8), set $u = 0$, $v = 4t$ and $w = 4$ (observe that $x$, $u$ and $v$ are even, so $2$ is a quintic residue modulo $p$); then $16p = 16t^4 + 50 \cdot 16t^2 + 125 \cdot 16$, and division by $16$ produces the polynomial $p = t^4 + 50t^2 + 125$. I'm sure that a suitable choice of the parameters will produce your polynomial.

  • $\begingroup$ I think you meant to write $w=4$ instead of $u=4$. $\endgroup$ – Yong Hao Ng Aug 14 '18 at 9:09
  • $\begingroup$ I do, thank you! $\endgroup$ – franz lemmermeyer Aug 14 '18 at 12:25
  • $\begingroup$ @franzlemmermeyer So coincidentally I found a case for $q=3$: Let $p=x^4 + 5x^3 + 10x^2 + 25$ be a prime. Then $3$ is a quintic residue $\pmod p$ if and only if $x=1 \pmod 3$. I'm not sure if this could be derived from the $(x, u, v, w)$ solution as stated in the article. $\endgroup$ – J. Linne Aug 15 '18 at 4:44

COMMENT.-It seems that this is a necessary and sufficient condition (hard to prove). Incidentally $f(x)=p$ for the three fist values $x=1,2,3$ for each of them there are five solutions. For $x=4,5,6,7$ $f(x)$ is not prime and there is no solution. For the next value $x=8$ one has $f(x)=p$ and again there are five solutions. For $f(9)$ and $f(10)$ both composite, there are no solution but for $f(11)=p$ immediately Wolfram gives five solutions again. Is this indicative of a pattern suggesting a necessary and sufficient condition?

  • 1
    $\begingroup$ So for $x=4,5,6,7$, $f(x)$ contains a prime factor $q$ such that $2$ is not a quintic residue $\pmod q$. Note that $q = 1\pmod 5$ (except for $5$, of course). The same, is true for $9, 10$. So, if my conjecture is true, this would explain why no solutions exists as you have observed. This is similar to the cubic reciprocity case first conjectured by Euler as I mentioned. $\endgroup$ – J. Linne Aug 12 '18 at 17:50
  • $\begingroup$ Nothing is said if it is not proved first. Also the sampling is too small to be a real indicator of a "trend". $\endgroup$ – Piquito Aug 12 '18 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.