# Quintic reciprocity conjecture

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.

UPDATE:

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.

• This looks hard. Has Euler's conjecture been resolved? – Robert Lewis Aug 12 '18 at 16:41
• @RobertLewis Yes, by Gauss if I recall correctly. – Lord Shark the Unknown Aug 12 '18 at 17:17
• @LordSharktheUnknown: I suspected that might be the case. Thanks! – 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.
• I think you meant to write $w=4$ instead of $u=4$. – Yong Hao Ng Aug 14 '18 at 9:09
• @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. – 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?
• 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. – J. Linne Aug 12 '18 at 17:50