16
$\begingroup$

David Cox's book Primes of The Form: $x^2+ny^2$ does a great job proving and motivating a lot of results for $n>0$. I was unable to find anything for negative numbers, let alone the case I am interested in, $n=-2$.

What is the reason for this? Maybe I am missing something. Any references to results involving primes of this form will be greatly appreciated.

$\endgroup$
1
  • 20
    $\begingroup$ Since $\mathbb{Z}[\sqrt2]$ is a UFD (hence a PID) you only need to understand for which odd primes the congruence $x^2-2y^2\equiv 0 \pmod p$ has a nontrivial solution, i.e. when $\left(\frac2p\right)=1$, that is when $p\equiv\pm1\pmod 8$. This leaves out 2, but then $2 = 2^2 - 2\cdot 1^2$.. $\endgroup$ Feb 19 '15 at 7:40
19
$\begingroup$

Take any square free $1 \neq n \in \mathbb{N}$ and recall that $R_n = \mathbb{Z}[\sqrt{n}]$ has a multiplicative norm function $N \colon R \to \mathbb{Z}$ given by $N(x + y\sqrt{n}) = x^2 -ny^2$ so a prime $p$ is of the required form iff $p$ is a norm in $R_n$ (i.e $p$ is in the image of $N$).

Now take $n$ which is not $1$ mod $4$ (this assures that $R_n$ is the ring of integers of $\mathbb{Q}(\sqrt{n})$) and for which the class number of $\mathbb{Q}(\sqrt{n})$ is $1$ (equivalently, $R_n$ is a principal ideal domain).

Clearly, $p = 2$ is of the required form for $n = 2$ ($x=2, y=1$), so assume $p$ is an odd prime number. Recall that $R_2$ is Euclidean with respect to $N$, so it is a principal ideal domain and $n = 2$ satisfies our assumptions.

Claim: The ideal $(p)$ splits completely (is a product two distinct prime ideals) in $R_n$ iff $p$ is of the required form.

Proof: We use the equivalent condition with the norm.

If $p = IJ$ is a splitting, taking norms of ideals we see that $N(IJ) = p^2$ so since the ideals are proper, $N(I) = p$. Since $R_n$ is principal, there exists some $a \in R_n$ such that $(a) = I$. Therefore, $N(a) = N(I) = p$, and $p$ is a norm as required.

If $p = N(x + y\sqrt{n})$, then $(p) = (x + y\sqrt{n})(x - y\sqrt{n})$ is the required splitting since the equality is obvious, and the primality of the factors follows from the fact that their norm is prime (equal to $p$). Furthermore the ideals are distinct as dividing a generator of one of them by a generator of the other does not belong to $R_n$.

Now, according to Neukirch’s Algebraic Number Theory proposition (8.5) $(p)$ splits completely iff $n$ is a quadratic residue mod $p$. For $n = 2$ this amounts to $p = \pm 1 \pmod 8$, by a lemma of Gauss.

$\endgroup$
4
  • $\begingroup$ Thanks!! I like this solution better, but I appreciate the variety of perspectives to solve this. $\endgroup$ Feb 20 '15 at 16:14
  • $\begingroup$ It is possible that I will write also a completely elementary solution soon. Hope it will be of interest. $\endgroup$
    – Pablo
    Feb 20 '15 at 16:19
  • 1
    $\begingroup$ How do you know that the norm of a is positive? You either need to prove it, or to show that there is a unit with negative norm (i.e., solution to $X^2-nY^2=-1$, for $n=2$ one may take $(1,1)$, of course...). $\endgroup$ Nov 15 '17 at 16:04
  • $\begingroup$ @LiorBary-Soroker you are right, I have completely missed this point. For instance, if $n=3$ and $p=11$ there seems to be a problem. As you correctly suggest, one needs the extra assumption that the negative Pell equation has a solution. $\endgroup$
    – Pablo
    Nov 15 '17 at 18:04
13
$\begingroup$

As $2 = x^2 -2y^2$ for $x = 2$ and $y=1$ we fix an odd prime number $p$, and

Claim: There exist integers $x,y$ such that $p = x^2 -2y^2$ if and only if $p \equiv \pm1 \mod 8$.

First if $p = x^2 -2y^2$ for some $x,y \in \mathbb{Z}$ then observing that the squares mod $8$ are $0,1,4$, we conclude that $p$ can only be $0,1,2,4,6,7$ mod $8$ but it is odd so $p \equiv \pm1 \mod 8$.

For the other direction, suppose that $p \equiv \pm1 \mod 8$. By a lemma of Gauss, there exists some $t \in (\mathbb{Z}/p\mathbb{Z})^*$ such that $t^2 \equiv 2 \mod p$. Define: $$S = \{a \in \mathbb{Z} \mid 0 \leq a \leq \sqrt{p}\}$$ and observe that $|S| > \sqrt{p}$. Consequently, $|S^2| > p$ so by the pigeonhole principle, there are different $(x_1,y_1),(x_2,y_2) \in S^2$ for which $x_1 - ty_1 \equiv x_2-ty_2 \mod p$, or equivalently, $x_1 - x_2 \equiv t(y_1 - y_2) \mod p$.

Set $x = |x_1 - x_2|, y = |y_1 - y_2|$ and note that $(x,y) \in S^2$ and that $x \equiv \pm t y \mod p$, so after squaring we get that $x^2 \equiv 2y^2 \mod p$, that is $p | x^2 - 2y^2$. On the other hand, $$|x^2 -2y^2| \leq |x|^2 +2|y|^2 < p + 2p = 3p.$$ Therefore, $x^2 -2y^2 \in \{-2p,-p,0,p,2p\}$.

Case 1: $x^2-2y^2 = 0$ which means that $x^2 = 2y^2$. Taking into account the number of times $2$ divides both sides (alternatively, set $x = 2x_0$ and apply infinite descent), we see that equality is possible if and only if $x = y = 0$, but this contradicts our assumption that $(x_1,y_1) \neq (x_2,y_2)$.

Case 2: $x^2 -2y^2 = 2p$. Clearly, $x$ is even. Set $a = x + y, b = x/2+y$ and observe that $a^2 -2b^2 = p$.

Case 3: $x^2 -2y^2 = -p$. Set $a = x + 2y, b = x + y$ and observe that $a^2 -2b^2 = p$.

Case 4: $x^2 -2y^2 = -2p$. Clearly, $x$ is even. Set $a = 2x + 3y, b =3x/2 + 2y$ and observe that $a^2 -2b^2 = p$.

Case 5: $x^2 -2y^2 = p$, which is what we want.

$\endgroup$
1
  • $\begingroup$ The part before the cases can be improved: first, you can actually get that $-2p<x^2-2y^2<p$ by doing cases on whether $x^2\geq 2y^2$ or $x^2<2y^2$. So you get $x^2-2y^2\in\{-p,0\}$. This means that you only need to check cases 1 and 3. $\endgroup$
    – dragoon
    Mar 8 at 12:54
4
$\begingroup$

Ummm. Just so you know, the same type of conclusion holds for $$ x^2 - p y^2 $$ for prime $p,$ $$ 5 \leq p \leq 197, \; \; \; \; p \equiv 1 \pmod 4. $$ For that matter, one may switch to the forms $$ x^2 + xy - \left( \frac{p-1}{4} \right) y^2 $$

For these forms, a number $n$ is represented if and only if $-n$ is represented. There is a solution to $x^2 - p y^2 = -1,$ a result in Mordell's book. Since every odd prime $q$ that satisfies $(p|q) = 1$ is represented by some form of the discriminant, and there is only one class of this discriminant, then all odd primes $q $ with $(q|p) = 1$ are integrally represented. Representation of the prime $2$ is a different matter, as we need $(2|p) = 1,$ so this works only when we further demand $p \equiv 1 \pmod 8.$

$\endgroup$

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.