A cubic equation, and integers of the form $a^2+192b^2$

Question

This question resembles my previous question A cubic equation, and integers of the form $a^2+32b^2$ , but seems to be more difficult.

We are trying to determine whether there are any integers $x,y,z$ such that $$ 1 - 12 x^2 - 4 y^2 + y z^2 = 0. \quad\quad\quad (1) $$ It is clear that $z$ is odd. After multiplying by $-16$ and rearranging, we can rewrite the equation as $$ (8y-z^2)^2 + 192x^2 = z^4 + 16. $$ Writing $z=2t+1$ and denoting $s=8y-z^2$, we obtain $$ s^2+192x^2=P(t), \quad\quad\quad (2) $$ where $P(t)=(2t+1)^4+16$. If this equation has no integer solutions, then so is the original one.

To solve this, we need to understand what integers are representable as $s^2+192x^2$. This seems non-trivial even for primes. Cox's famous book "primes of the form $x^2+n y^2$" gives a characterization in terms of certain polynomial $P_n$, but contains no tables of $P_n$ for small $n$, so what is $P_{192}$ ?

In fact, I cannot even find online a list of reduced quadratic forms by discriminant long enough to contain $D=-4 \cdot 192 = -768$. In particular, what is the class number $h(-768)$?

But of course the main question is: are there any integers $x,y,z$ satisfying (1)?

Answer 1

Some partial information from hand calculation:

The class number for discriminant $-768$ is 8 with the reduced forms $[1,0,192]$, $[3,0,64]$, $[4,4,49]$, $[7,\pm4,28]$, $[13,\pm8,16]$, and $[12,12,19]$. The class group is $C_2\times C_4$, the product of cyclic groups of orders 2 and 4. There are four genera, with two forms in each genus. The principal genus consists of $[1,0,192]$ and $[4,4,49]$ since $[4,4,49]$ is the square of $[7,4,28]$, as $[7,4,28]$ represents 7 and $[4,4,49]$ represents $7^2=49$. The problem is therefore to determine which numbers are represented by $[1,0,192]$ and which are represented by $[4,4,49]$.

Answer 2

According to pari/gp, there are no solutions to (2) with $0 < t \le 10^5$.

pari/gp code:

{
test1()=
Q1=Qfb([1,0,192]);
for(T=1,10^5,
A=(2*T+1)^4+16;so=qfbsolve(Q1,A);
if(so !=[],print([T,so])));
}
test1()

Answer 3

Ups, it looks like the equation is easy but I mislead the readers by wrong idea.

It is clear that $y>0$. Modulo $3$ analysis shows that $y$ is not divisible by $3$, while modulo $4$ analysis shows that $y=3$ modulo $4$. Hence, $y$ is either $7$ or $11$ modulo $12$. Now rewrite the equation as $$ -12x^2+yz^2 = 4y^2-1=(2y-1)(2y+1). $$ and consider it as quadratic in $x$ and $z$ with parameter $y$. The discriminant of the quadratic form on the left-hand side is $D=48y$. If $y=7$ modulo $12$, then the equation implies that $\left(\frac{48y}{2y-1}\right)=1$, but computation shows that it is $-1$. Similarly, if $y=11$ modulo $12$, then the equation implies that $\left(\frac{48y}{2y+1}\right)=1$, but computation shows that it is $-1$.

Answer 4

I will use $n,D$ for the special values of interest only, $$ \bbox[lightyellow]{\qquad \begin{aligned} n &= 192\ ,\\ D &= -4n=-768\ . \end{aligned}\qquad} $$ Following the reference ([Cox] below)

Cox, Primes of the form $x^2+ny^2$ [...] Theorem 9.2, §9.B and Examples in §9.C,

we are dealing with the field the order $$ \begin{aligned} K &=\Bbb Q(a)=\Bbb Q(\sqrt{-3})\ ,\qquad a=\sqrt{-3}\ , \\ &\qquad\text{ with integer basis $1$ and $w=\frac 12(1+a)$, and its order} \\[2mm] \mathcal O &=\Bbb Z[\sqrt{-n}]\\ &=\Bbb Z\cdot 1 +\Bbb Z\cdot 8a=\Bbb Z\cdot 1 +\Bbb Z\cdot 16w \end{aligned} $$ of discriminant $D=-4n=-768$ and conductor $16$. The ideal class group of the order $\mathcal O$ has the structure $$ C(\mathcal O)\cong \Bbb Z/4\oplus\Bbb Z/2\ , $$ as pari/gp shows it below. For the convenience of the reader, here is the adapted version of this theorem for our case.

Theorem 9.2 [Cox] and the particular case $\bf n=192$:

Let $n>0$ be an integer. There is a monic irreducible polynomial $f_n\in\Bbb Z[x]$ of degree $h(-4n)$ such that if an odd prime $p$ divides neither $n$ nor the discriminant of $f_n$ then: $$ p = x^2+ny^2\Longleftrightarrow \left\{ \begin{aligned} &\left(\frac {-n}p\right)=1\text{ and}\\ &f_n(x)\equiv 0\text{ modulo $p$ has an integer solution.} \end{aligned} \right. $$ Furthermore, $f_n(x)$ may be taken to be the minimal polnomial of a real algebraic integer $\alpha$ for which $L=K(\alpha)$ is the ring class field of the order $\mathcal O=\Bbb Z[\sqrt{-n}]$ in the imaginary quadratic field $K=\Bbb Q(\sqrt{-n})$.

Finally, if $f_n$ is any monic integer polynomial of degree $h(-4n)$ for which the above equivalence holds, then $f_n(x)$ is irreducible over $\Bbb Z$ and is the minimal polynomial of a primitive element of the ring class field $L$ described above.

---

For the special case $n=192$ we can take $$ \bbox[lightyellow]{ \qquad f_n(x) = (x^2-2)^4+9\ , \qquad} $$ so $L$ can be taken to be the number field generated over $K$ by a root of this polynomial.

---

Some details follow now, parallel to the many chapters in [Cox].

Investigation of the binary quadratic forms with the given discriminant, here aided by pari/gp, see also

https://pari.math.u-bordeaux.fr/pub/pari/manuals/2.3.5/tutorial.pdf , §10 , pages 27 - 31 .

the code fragments is just a quick way to type elements, and may be also useful in similar situation:

? n = 192;
? D = -4*n;
? quadclassunit(D, , [1, 6])
%56 = [8, [4, 2], [Qfb(7, 4, 28), Qfb(12, 12, 19)], 1]

(The order of the class group is $8$, the invariant factors are $4,2$, representing generators are the binary quadratic forms $[a,b,c]=[ax^2+bxy+cy^2]$ given by $[7,4,28]$ and $[12,12, 19]$.) To see them all, we may want compute:

? f = Qfb(7, 4, 28); g = Qfb(12, 12, 19);
? for(j=0, 3, for(k=0, 1, print("f^", j, " * g^", k, " = ", f^j * g^k))) 

f^0 * g^0 = Qfb(1, 0, 192)
f^0 * g^1 = Qfb(12, 12, 19)
f^1 * g^0 = Qfb(7, 4, 28)
f^1 * g^1 = Qfb(13, -8, 16)
f^2 * g^0 = Qfb(4, 4, 49)
f^2 * g^1 = Qfb(3, 0, 64)
f^3 * g^0 = Qfb(7, -4, 28)
f^3 * g^1 = Qfb(13, 8, 16)

? \\ also note:
? f^4
%59 = Qfb(1, 0, 192)
? g^2
%60 = Qfb(1, 0, 192)

The $(j,k)$-loop shows all elements $f^jg^k$ in the class group, $f, g$ being the generating binary quadratic forms (as classes) $f=[7,4,28]$, and $g=[12,12,19]$. They have orders $4,2$.

The neutral / principal element is $[1, 0, 192]=[1,0,n]$. It is a square, and the only other square is $f^2=[4,4,49]$. This is the genus containing the principal form, $C(\mathcal O)^2$.

(Alternatively for the class number only we may ask for qfbclassno(D) and obtain the 8.)

---

---

Let us collect some experimental data now. For $t=4$ the value $P(t)=(8+1)^4+16=6577$ is by chance a prime number.

We are searching for an integer solution/representation
$s^2+192x^2=6577$ for it.

There is no such solution in integers, concluded after a small $x$-loop, but there is one representation when we allow the denominator $2$ for $x$, i.e. for the related quadratic form $(S, X)\to S^2 + 48X^2$. We have $(s_0,x_0)=(65,7/2)$, $(S_0,X_0)=(65, 7)$.

However, we have an integer representation for the other form in the principal genus, $4s^2+4sx+49x^2=6577$, it is $(s,x)=(29,7)$. Similar examples can be given also for other values, it is maybe useful to collect this experimental data in a table for a first orientation: $$ \tiny \begin{array}{|r|r|c|c|c|} \hline t & P(t)& & s^2 + 192x^2=P(t)& 4s^2 + 4sx + 49x^2=P(t)\\\hline 1 & 97 & \text{PRIME} & (5/2,\ 11/16) & (3,\ 1)\\\hline 4 & 6577 & \text{PRIME} & (65,\ 7/2) & (29,\ 7)\\\hline 10 & 17^{2} \cdot 673 & & (425,\ 17/2) & (221,\ 17)\\\hline 13 & 531457 & \text{PRIME} & (247,\ 99/2) & (74,\ 99)\\\hline 22 & 4100641 & \text{PRIME} & (2023,\ 13/2) & (1005,\ 13)\\\hline 25 & 6765217 & \text{PRIME} & (295,\ 373/2) & (334,\ 373)\\\hline 28 & 10556017 & \text{PRIME} & (3103,\ 139/2) & (1482,\ 139)\\\hline 31 & 313 \cdot 50329 & & (1265,\ 543/2) & (1839,\ 301)\\\hline 37 & 4993 \cdot 6337 & & (3607,\ 623/2) & (1492,\ 623)\\\hline 43 & 57289777 & \text{PRIME} & (7553,\ 71/2) & (3812,\ 71)\\\hline 46 & 73 \cdot 1024729 & & (7495,\ 623/2) & (4059,\ 623)\\\hline 52 & 121550641 & \text{PRIME} & (8129,\ 1075/2) & (4602,\ 1075)\\\hline 55 & 2521 \cdot 60217 & & (3857,\ 1689/2) & (2773,\ 1689)\\\hline 70 & 395254177 & \text{PRIME} & (9497,\ 2521/2) & (6009,\ 2521)\\\hline 73 & 466948897 & \text{PRIME} & (21415,\ 417/2) & (10916,\ 417)\\\hline 79 & 639128977 & \text{PRIME} & (24655,\ 807/2) & (11924,\ 807)\\\hline 94 & 1275989857 & \text{PRIME} & (34745,\ 1197/2) & (16774,\ 1197)\\\hline 97 & 3433 \cdot 421177 & & (31673,\ 3037/2) & (14318,\ 3037)\\\hline 103 & 1836036817 & \text{PRIME} & (42833,\ 169/2) & (21501,\ 169)\\\hline 127 & 1129 \cdot 3745129 & & (60943,\ 3273/2) & (25073,\ 7025)\\\hline 130 & 4640470657 & \text{PRIME} & (36727,\ 8281/2) & (22504,\ 8281)\\\hline 133 & 26713 \cdot 190249 & & (23767,\ 9701/2) & (33844,\ 4863)\\\hline 139 & 193 \cdot 31394929 & & (2623,\ 11229/2) & (9871,\ 11167)\\\hline 160 & 73 \cdot 145444489 & & (71503,\ 10709/2) & (41106,\ 10709)\\\hline 166 & 27457 \cdot 447841 & & (92455,\ 8837/2) & (50646,\ 8837)\\\hline 178 & 97 \cdot 9241 \cdot 18121 & & (117065,\ 7273/2) & (17386,\ 16821)\\\hline 184 & 18539817937 & \text{PRIME} & (48593,\ 18359/2) & (33476,\ 18359)\\\hline 190 & 141793 \cdot 148609 & & (137095,\ 6887/2) & (20458,\ 20749)\\\hline 199 & 25344958417 & \text{PRIME} & (85615,\ 19373/2) & (52494,\ 19373)\\\hline 202 & 26904200641 & \text{PRIME} & (131479,\ 14155/2) & (72817,\ 14155)\\\hline 205 & 28534304257 & \text{PRIME} & (161993,\ 6911/2) & (84452,\ 6911)\\\hline 223 & 39923636497 & \text{PRIME} & (101297,\ 24859/2) & (38219,\ 24859)\\\hline 226 & 42110733697 & \text{PRIME} & (129463,\ 22981/2) & (53241,\ 22981)\\\hline 232 & 41281 \cdot 1132561 & & (117137,\ 26233/2) & (107855,\ 9293)\\\hline 235 & 409 \cdot 120326233 & & (180223,\ 18671/2) & (99447,\ 18671)\\\hline 241 & 601 \cdot 90555337 & & (226375,\ 8137/2) & (109119,\ 8137)\\\hline 247 & 60037250641 & \text{PRIME} & (226577,\ 13463/2) & (106557,\ 13463)\\\hline 253 & 337 \cdot 196065841 & & (150583,\ 30069/2) & (90326,\ 30069)\\\hline 256 & 226777 \cdot 305401 & & (112975,\ 34307/2) & (101312,\ 20209)\\\hline 268 & 82129 \cdot 1012513 & & (160735,\ 34557/2) & (63089,\ 34557)\\\hline 277 & 73 \cdot 1299717817 & & (61207,\ 43573/2) & (52390,\ 43573)\\\hline 286 & 1249 \cdot 86308993 & & (241913,\ 32041/2) & (162548,\ 15969)\\\hline \end{array} $$ So all the time the odd(est) prime two is in the way of $[1,0,192]$, and always $[4,4,49]$ is winning.

---

---

For the computation of $f_n$ i followed the receipt in §9, §11 in loc. cit searching for subfields dihedral Galois structure. So far, here are some polynomials of degree $8$ that experimentally work up to some higher bound for tested prime numbers: $$ \begin{aligned} & x^{8} - 8 x^{7} - 56 x^{6} + 64 x^{5} + 2592 x^{4} + 11840 x^{3} + 30400 x^{2} + 32000 x + 40000\\ & x^{8} - 48 x^{7} + 736 x^{6} - 1408 x^{5} + 7232 x^{4} - 15872 x^{3} + 27648 x^{2} - 20480 x + 25600\\ & x^{8} - 8 x^{7} - 24 x^{6} + 160 x^{5} + 800 x^{4} - 832 x^{3} + 64 x^{2} + 64\\ & x^{8} - 8 x^{7} - 24 x^{6} + 352 x^{5} - 736 x^{4} - 2368 x^{3} + 16960 x^{2} - 38400 x + 40000\\ & x^{8} - 8 x^{7} - 72 x^{6} + 448 x^{5} + 2000 x^{4} - 3712 x^{3} - 23552 x^{2} - 58368 x - 40064\\ & x^{8} - 8 x^{7} - 24 x^{6} + 160 x^{5} + 880 x^{4} - 3072 x^{3} + 1920 x^{2} - 27648 x + 67968\\ & x^{8} - 8 x^{7} + 8 x^{6} + 96 x^{5} - 208 x^{4} + 1664 x^{2} - 2048 x + 1408\\ \end{aligned} $$

The third polynomial in the list looks nice. We let sage give us its subfields of degree four, take a defining polynomial, and let this polynomial be factorized over $\Bbb Q(i)$. (Here, $i$ was a choice after looking at the subfields of degree two.) It turns out that a factor like $(x^2 - 2(1+4i)x - 39 + 8i$ shows up. Its (half)discriminant is $(1+4i)^2+39-8i=24=3i(1+i)^6$. So we may want to adjoin $\sqrt{3i}$, which leads to further adjoining $b=\sqrt 6$. We let than factor the third polynomial in the list over $\Bbb Q(i,b)$, and the shown factors lead to taking the root of $8-2b-2ib$. And half of $\sqrt(8-2b-2ib)$ has the minimal polynomial

sage: (1/2*sqrt(8 - 2*sqrt(6) - 2*sqrt(-6))).minpoly()
x^8 - 8*x^6 + 24*x^4 - 32*x^2 + 25

which is $(x^2-2)^4+9$.

---

---

The table was genarated by the sage code:

P = lambda t: (2*t + 1)^4 + 16
Q = DiagonalQuadraticForm(QQ, [1, 192])
R = QuadraticForm(QQ, 2, [4, 4, 49])

for t in [0..300]:
    try:
        pt = P(t)
        info = "\\text{PRIME}" if pt.is_prime() else ""
        sQ, xQ = Q.solve(pt)
        sR, xR = R.solve(pt)
        sQ, xQ, sR, xR = abs(sQ), abs(xQ), abs(sR), abs(xR)
        print(f"{t:>4} & {latex(pt.factor())} & "
              f"{info}"
              f" & ({sQ},\\ {xQ}) & ({sR},\\ {xR})\\\\\\hline")
    except ArithmeticError:
        pass

---

The code for the polynomials of degree $8$ above is:

x^8 - 8*x^7 - 56*x^6 + 64*x^5 + 2592*x^4 + 11840*x^3 + 30400*x^2 + 32000*x + 40000
x^8 - 48*x^7 + 736*x^6 - 1408*x^5 + 7232*x^4 - 15872*x^3 + 27648*x^2 - 20480*x + 25600
x^8 - 8*x^7 - 24*x^6 + 160*x^5 + 800*x^4 - 832*x^3 + 64*x^2 + 64
x^8 - 8*x^7 - 24*x^6 + 352*x^5 - 736*x^4 - 2368*x^3 + 16960*x^2 - 38400*x + 40000
x^8 - 8*x^7 - 72*x^6 + 448*x^5 + 2000*x^4 - 3712*x^3 - 23552*x^2 - 58368*x - 40064
x^8 - 8*x^7 - 24*x^6 + 160*x^5 + 880*x^4 - 3072*x^3 + 1920*x^2 - 27648*x + 67968
x^8 - 8*x^7 + 8*x^6 + 96*x^5 - 208*x^4 + 1664*x^2 - 2048*x + 1408

They were obtained by asking for the subfields of degree eight of the ray class field obtained via

`pari.quadray(-3, 16)`

To put the hands on the subfields, one may try

F.<y>  = CyclotomicField(6)
RF.<X> = PolynomialRing(F)
f      = pari.quadray(-3, 16)
N.<v>  = NumberField(RF(f))

And then we use Theorem 9.2 again to rule out some of the subields of degree eight, that do not work for our $n$.

---

Here is a computer check for primes up to some bound that the claimed $f_n$ works.

R.<X> = PolynomialRing(ZZ)
f = (X^2 - 2)^4 + 9
q = BinaryQF([1, 0, 192])


def check(bound=10^8, verbose=False):
    ok = True
    for p in primes(bound):
        sols = Q.solve_integer(p)
        kron = bool(kronecker(-n, p) == 1)
        roots = f.base_extend(GF(p)).roots(GF(p), multiplicities=0)
        if verbose:
            print(f"p = {p:>3} | {sols} | {kron:<5} | {roots}")
        ok = bool(sols) == bool(kron and roots)
        if not ok:    break
    return ok

Then the check up to the default bound is:

sage: check()
True

A verbose check with a lower bound may be started with check(bound=1000, verbose=True).