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I already posted this question on MSE.

Using theorem $IV$ from this article, it is possible to prove that when $p$ is a prime such that $p ≡ 3\bmod4$, $x ≢ y\bmod{p}$ and $\gcd(x,y) = 1$, then the equation $$ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $$ ever admits solutions for some integers $\{a, b\}$ (see proof below).

I want to know how $\{a, b\}$ can be expressed as a function of $\{x, y\}$: the case $p=3$ is easy as $$ \begin{align} a &= \dfrac{x-y}{2}\\ b &= \dfrac{x+y}{2} \end{align} $$ (of course not all the solutions have this form). With the great help of Will Jagy on MSE, the closed form for the following cases are $$ \begin{align} p &= 7\\ a &= \frac{ (x-y) (2x^2 + 3xy + 2y^2 )}{ 2},\\ b &= \frac{xy(x+y)}{2},\\ p &= 11\\ a &= \frac{( x-y) (2x^4 + 3x^3 y + x^2 y^2 + 3 x y^3 + 2 y^4)}{ 2},\\ b &= \frac{ xy(x+y)(x^2 - xy+y^2)}{ 2}\\ p &= 19\\ a &= \frac{ (x-y) (2x^8 + 3x^7y - x^6y^2 + 2x^5y^3 + 7x^4y^4 + 2x^3y^5 - x^2y^6 + 3xy^7 + 2y^8 )}{ 2},\\ b &= \frac{ x y (x +y) ( x^2 -xy +y^2)(x^4 - x^2 y^2 + y^4 )}{ 2} \end{align} $$ What can be said in general?

I guess it is possible to restate the problem in terms of cyclotomic binary forms: if we define $$ f(k) = y^{\phi(k)} \Phi_k(x/y) $$ where $ \phi(k)$ is Euler's phi function and $ \Phi_k(x/y) $ is the $k$-th cyclotomic polynomial, so that $$ f(p) = \dfrac{x^p - y^p}{x - y} $$ and $$ \begin{align} f(7) &= a^2 + 7\left (\dfrac{xyf(2)}{2} \right )^2\\ f(11) &= a^2 + 11\left (\dfrac{xyf(2)f(6)}{2} \right )^2\\ f(19) &= a^2 + 19\left (\dfrac{xyf(2)f(6)f(12)}{2} \right )^2 \end{align} $$

Proof: when $(x - y) ≢ {0}\bmod{p}$ then we have that $$ \dfrac{x^p - y^p}{x - y} = P(p) $$ where $P(p)$ is the arithmetic primitive factor of $ (x^p - y^p) $ defined in that article (p. $175$). With other considerations we can say that each one of those primitive prime factors is of the form $(2k_i p+1)$ where $k_i$ $∈ N$.
Now $q\mid{a^2+pb^2} $, with $q\nmid p$, iff $ \left(\frac{-p}{q}\right) = 1$ and equivalently $$ q ≡ \{q_1, q_2, q_3,\ldots\}\bmod{4p} $$ (see corollary $1.19$ in David Cox's book Primes of the Form $x^2+ny^2$), so being $q=2k_i p+1$, we have just $q_1=1$ and $q_2 = 2p+1$. Using that $p ≡ 3\bmod4$, we have $$ \left(\frac{-p}{2p+1}\right) = 1 $$ and so $$ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $$ ever admits solutions for some integers $\{a, b\}$.

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    $\begingroup$ as I had indicated, the fact that a prime $q \equiv 1 \pmod p$ is no guarantee of a representation as $q = u^2 + p v^2$ or as $q = s^2 + st + \frac{1+p}{4} t^2,$ the parent form. So I don't see your last paragraph as a proof of your conjecture. For $p=23$ the two binary forms agree for this purpose. I've got a few dozen primes expressible as your $q =\frac{x^{23} - y^{23}}{x-y}.$ By factoring $x^3 - x + 1 \pmod q$ I can tell whether $q = u^2 + 23 v^2.$ So far, all succeed. See zakuski.math.utsa.edu/~jagy/Hudson_Williams_1991.pdf $\endgroup$
    – Will Jagy
    Commented Jun 26 at 16:58
  • $\begingroup$ The thing about 7, 11, 19 is class number one. en.wikipedia.org/wiki/Heegner_number $\endgroup$
    – Will Jagy
    Commented Jun 26 at 17:59
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    $\begingroup$ @DanielAsimov it is an older usage of the word ever, now largely obsolete. I looked up the etymology, it all makes sense... I would have written "always," or perhaps just omitted the word ever. $\endgroup$
    – Will Jagy
    Commented Jun 26 at 18:06
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    $\begingroup$ Andre Weil, in Number Theory: An approach through History, on pages 336 and 337, says that Gauss proved this. Calling $F= \frac{x^p - y^p}{x-y}$ splits into two factors in $\mathbb Q(\sqrt {\pm p}) ,$ leading to polynomials $P,Q$ with half-integer coefficients and $F = P^2 - p Q^2$ when $p \equiv 1 \pmod 4,$ but $F = P^2 + p Q^2$ when $p \equiv 3 \pmod 4,$ $\endgroup$
    – Will Jagy
    Commented Jun 29 at 1:11
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    $\begingroup$ it's in the Disquisitiones, articles 124 and 357. In my translation pages 82 and 439-440. $\endgroup$
    – Will Jagy
    Commented Jun 29 at 1:44

2 Answers 2

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Values for $a$ and $b$ as polynomials in $x,y$, when they exist, correspond to factorization of $\frac{x^p+y^2}{x+y}$ over the imaginary field $K_p:=\mathbb Q(\sqrt{-p})$ of the form: $$\frac{x^p+y^p}{x+y} = (a(x,y)+\sqrt{-p}b(x,y))\cdot (a(x,y)-\sqrt{-p}b(x,y)).$$ For $p\equiv1\pmod4$, we can similarly work in $K_p:=\mathbb Q(\sqrt{p})$ and look for factorization of the form: $$\frac{x^p+y^p}{x+y} = (a(x,y)+\sqrt{p}b(x,y))\cdot (a(x,y)-\sqrt{p}b(x,y)).$$

This Sage code (updated 2024-06-30) factors $\frac{x^p+1}{x+1}$ over $K_p$ and converts the result into the required form, and reports the (univariate) polynomials $a(x,1)$ and $b(x,1)$. As an example, it reports $a$ and $b$ for all primes below $100$.

Example for $p=23$: $$a = (x + 1) \cdot (x^{10} - \frac{3}{2} x^{9} - x^{8} + 5 x^{7} - \frac{17}{2} x^{6} + \frac{21}{2} x^{5} - \frac{17}{2} x^{4} + 5 x^{3} - x^{2} - \frac{3}{2} x + 1),$$ $$b= \frac{1}{2} \cdot (x - 1) \cdot x \cdot (x^{8} + x^{5} - x^{4} + x^{3} + 1).$$

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    $\begingroup$ Andre Weil, in Number Theory: An approach through History, on pages 336 and 337, says that Gauss proved this. Calling $F= \frac{x^p - y^p}{x-y}$ splits into two factors in $\mathbb Q(\sqrt {\pm p}) ,$ leading to polynomials $P,Q$ with half-integer coefficients and $F = P^2 - p Q^2$ when $p \equiv 1 \pmod 4,$ but $F = P^2 + p Q^2$ when $p \equiv 3 \pmod 4,$ $\endgroup$
    – Will Jagy
    Commented Jun 29 at 1:11
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    $\begingroup$ in my English translation of the Disquisitiones Gauss announces the result in article 124, page 82, then gives detail in article 357, pages 439-440 $\endgroup$
    – Will Jagy
    Commented Jun 29 at 1:28
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    $\begingroup$ @MaxAlekseyev Thank you so much! I tried with a few other cases, but it seems that $b$ is not factorizable in terms of cyclotomic polynomials, like I showed for the cases $p=7, 11, 19$, that just happened by chance or is there an underlying reason? Thank you! $\endgroup$
    – user967210
    Commented Jun 29 at 8:39
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    $\begingroup$ @WillJagy: Nice find! I suspected that this subject should be well studied. $\endgroup$
    – Max Alekseyev
    Commented Jun 29 at 21:35
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    $\begingroup$ @user967210: I think it was just by chance. The factors of $b$, besides $x-1$, are generally palindromic but do not necessarily divide $\Phi_p$. $\endgroup$
    – Max Alekseyev
    Commented Jun 29 at 22:05
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Friday, June 28. I found a nice exposition by David Savitt

https://pi.math.cornell.edu/~web401/steve.gauss17gon.pdf

from which this is page 32

David A. Cox, in Galois Theory, gives an account of Gauss's theory of periods
http://zakuski.math.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf

Cox does only a few examples, but Reuschle filled a book with nothing but.

https://archive.org/details/tafelncomplexer00unkngoog/page/n7/mode/2up

enter image description here

The first few cases where $ z = \frac{x^{23} - y^{23}}{x-y} $ turned out to be prime. I wrote some software to solve $z = u^2 + 23 v^2$ in a reasonable time.

x: 2 y: -1 z: 2796203  z prime ? 2   
990^2 + 23 * 281^2 =  2796203 

x: 3 y: -1 z: 23535794707  z prime ? 2  
 118750^2 + 23 * 20253^2  = 23535794707

x: 5 y: 3 z: 5960417405949649  z prime ? 1   ?
  38872207^2 + 23 * 13908660^2   = 5960417405949649

x: 6 y: 5 z: 777809294098524691  z prime ? 1 ? 
   827130254^2  + 23 * 63815235^2   = 777809294098524691

x: 7 y: -1 z: 3421093417510114543  z prime ? 1?
   1755643660^2 + 23 *  121370571^2   = 3421093417510114543

x: 7 y: -2 z: 3040971926676589439  z prime ? 1? 
 1438562808^2  + 23 *  205522555^2   = 3040971926676589439

x: 10 y: 1 z: 11111111111111111111111  z prime ? 1    ?
 102063239244^2 + 23 * 5493895055^2   = 11111111111111111111111
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  • $\begingroup$ Wow! Thank you so much for the reference! $\endgroup$
    – user967210
    Commented Jun 29 at 8:08

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