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The current question is motivated by this question. It is known that the number of imaginary quadratic fields of class number 3 is finite. Assuming the answer to this question is affirmative, I came up with the following question.

Let $f(X) = X^3 + aX + b$ be an irreducible polynomial in $\mathbb{Z}[X]$. Let $p = -(4a^3 + 27b^2)$ be the discriminant of $f(X)$. We consider the following conditions.

(1) $p = -(4a^3 + 27b^2)$ is a prime number.

(2) The class number of $\mathbb{Q}(\sqrt{p})$ is 3.

(3) $f(X) \equiv (X - s)^2(X - t)$ (mod $p$), where $s$ and $t$ are distinct rational integers mod $p$.

Question Are there infinitely many primes $p$ satisfying (1), (2), (3)?

If this is too difficult, is there any such $p$?

I hope that someone would search for such primes using a computer.

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I believe it is unknown whether there are infnitely many primes satisfying your condition (2). – Gerry Myerson Aug 22 '12 at 1:42
@GerryMyerson I guess there are many(if not infinitely many) primes satisfying the condition (2). I'd like to know, at least, if there exists any prime number satisfying (1), (2), (3). – Makoto Kato Aug 22 '12 at 7:24
So, how many primes have you looked at? – Gerry Myerson Aug 22 '12 at 12:58
@GerryMyerson I looked at the following link. Hilbert class fields of all real quadratic fields of discriminants up to 30000 are given. I found (p, a, b) = (229, -4, -1), (1373, -8, -5), (2713, -13, -15). Each of these satisfies (1) and (2), but I don't know if they satisy (3). – Makoto Kato Aug 22 '12 at 17:35
How many elliptic curves $y^2 = x^3 + ax + b$ of discriminant $2^4*p$ are there with multiplicative reduction at p? The quadratic twist of this elliptic curve by its discriminant might also be relevant. – Thom Tyrrell Oct 5 '12 at 14:25
up vote 3 down vote accepted

For (229, -4,-1) the polynomial factors as $(x-200)^2(x-58)$

For (1373, -8,-5) the polynomial factors as $(x-860)(x-943)^2$

For (2713, -13,-15) the polynomial factors as $(x-520)^2(x-1673)$

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I take it $(229,-4,-1)$ means $p=229$, $a=-4$, $b=-1$, and I take it you are claiming the class number of ${\bf Q}(\sqrt{229})$ is 3. – Gerry Myerson Oct 5 '12 at 0:34
@GerryMyerson See Kato's remarks above. – i. m. soloveichik Oct 5 '12 at 13:10
Ah.${}{}{}{}{}$ – Gerry Myerson Oct 5 '12 at 13:24
Thanks. May I ask how you found the factorizations? – Makoto Kato Oct 5 '12 at 22:23
I wrote a Maple program. – i. m. soloveichik Oct 5 '12 at 22:34

Here's some code I wrote using Sage:

def quadClassNumber(p):
    K.<alpha> = NumberField([x^2-p]);
    return K.class_number();  

for p in Primes():
    if p < U and quadClassNumber(p)==3:
        J = EllipticCurve([0,-2^4*3^3*p]);
        L = J.integral_points();
        if len(L) > 0:
            for (a,b,1) in L:
                A = a/(-2^2*3);
                B = b/(2^2*3^3);
                if A.is_integral() and B.is_integral():
                    E = EllipticCurve([A,B]);
                    if E.has_multiplicative_reduction(p):
                        print "p=",p;
                        print E;
                        print E.discriminant().factor();
                        R.<t> = PolynomialRing(IntegerModRing(p),1);
                        print R(E(0,1,0).division_points(2,true)/4).factor();
                        print "    ";
    elif p > U:

The integral points of the elliptic curve J correspond (roughly) to pairs of integers $(A,B)$ satisfying $p = -(4A^3 + 27B^2)$. J is an integral model of the elliptic curve associated with this equation, but the use of J and Sage's integral_points() function comes at the cost of some powers of 2 and 3 that appear in the definitions of $A$ and $B$.

I searched for primes up to 10,000. For instance, with p=8581, a = -16 and b=17 work, and the associated polynomial factors mod 8581 as $(x + 6166)^2(x + 4830)$.

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