As the title suggests, I'm trying to prove that there are no solutions to $a^2 - 223b^2 = -3$ (with $a,b\in \mathbb{Z}$). Ordinarily, taking both sides $\mod n$ for some clever choice of $n$ proves that no solutions exist, but I've tried this method for $n = 2$ through $18$, and all have solutions. Moreover, we have that \begin{align} \left(\frac{-3}{223} \right) &= \left(\frac{220}{223} \right)=\left(\frac{55}{223} \right) = \left(\frac{11}{223}\right)\left(\frac{5}{223}\right) \\ &= -\left(\frac{223}{11}\right)\left(\frac{223}{5}\right) = -\left(\frac{3}{11}\right)\left(\frac{3}{5}\right) \\ &= -(1)(-1) = 1, \end{align} so $-3$ is a square $\mod 233$, so Legendre symbols aren't much of a help.

I suppose if I keeping taking both sides $\mod n$ for increasingly large $n$, then I'll eventually get the answer, but there must be a more intelligent way to do this. Any help is appreciated.

As a bit of background, this came up in a proof that the ideal class group of $\mathbb{Z}[\sqrt{223}]$ is $\mathbb{Z}/3\mathbb{Z}$. In this post, the OP casually states that there is no element of norm $-3$, and this is not an obvious fact to me.

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    $\begingroup$ If you work $mod(3)$ we get the equation $a^2=2b^2\;mod(3)$ which tells us that $a=0=b\;mod(3)$. But then the left hand is divisible by $9$ and the right hand isn't. $\endgroup$ – lulu Aug 3 '15 at 19:05
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    $\begingroup$ Don't we have $223 \equiv 1 \pmod{3}$ rather than $2$? Then we have $a^2 \equiv b^2 \pmod{3}$ which doesn't lead anywhere. $\endgroup$ – Marcus M Aug 3 '15 at 19:08
  • $\begingroup$ Ah! I had transcribed it as $233$. Different problem. Shouldn't work without my glasses. Sorry! $\endgroup$ – lulu Aug 3 '15 at 19:10
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    $\begingroup$ I feel your pain, and did exactly that multiple times. $\endgroup$ – Marcus M Aug 3 '15 at 19:10
  • $\begingroup$ math.stackexchange.com/questions/78863/a-diophantine-equation $\endgroup$ – mathlove Aug 3 '15 at 19:34

Equivalently, there should be no solutions to

223a^2 - b^2 = 3

The binary quadratic form 223x^2 - y^2 has discriminant 892. Because 892 is a square modulo 3, there is some form of discriminant 892 that represents 3. However, it turns out that there are some forms of discriminant 892 that represent 3 and some that don't. Distinguishing the two will take you much deeper than just considering congruences or the Legendre symbol and into the theory of binary quadratic forms. If you don't know much of this theory, I highly recommend the book "Primes of the form x^2 + ny^2" by Cox.

Using Zagier's theory of reduction, we can give a reasonably simple check that 3 cannot be represented by 223x^2 - y^2. Zagier proves in his 1981 book "Zetafunktionen und quadratische körper" that every representation $n= 223x^2 - y^2$ will be equivalent to a unique representation $n = f(u,v)$ where $f$ has the form $Ax^2 + Bxy + Cy^2$ with $A > 0$, $C > 0$, and $B > A+C$; $f$ is in the same equivalence class as $223x^2 - y^2$; and $u$ and $v$ are integers with $u > 0$ and $v \geq 0$. Using Keith Matthew's tool at:


or working by hand, one can compute that there are 41 Zagier-reduced forms in the equivalence class of 233x^2 - y^2, and the smallest coefficient appearing is 27. Clearly, there is no way to represent 3 by such a form using nonnegative integers $u$ and $v$, so no such representation exists.

(Incidentally, a form of discriminant 892 that does represent 3 is, for instance, $3x^2 + 26xy - 18 y^2$.)

Edit: in response to the motivation you supplied in the edit to your post:

You can prove using binary quadratic forms that the class group of $K = \mathbb{Q}(\sqrt{223})$ is $\mathbb{Z}/3\mathbb{Z}$ without reference to the Minkowski bound. One can compute the equivalence classes of forms of discriminant 892 (the discriminant of $K$) by standard methods and find that there are 6 classes. The class group of forms is isomorphic to the narrow class group of $K$, which is either the same size as the class group or double the size. They are the same size if and only if the fundamental unit of $K$ has norm $-1$.

A computation shows the fundamental unit is $224+15\sqrt{223}$, which has norm 1, so the class group of $K$ is half the size of the group of forms. Thus, it has order $3$.


Looking at the continued fraction for $\sqrt{223}$, this is pretty clear.

The continued fraction for $\sqrt{223}$ is $$ (14;\overbrace{1,13,1,28}^{\large\text{repeat}}) $$ If $223q^2-p^2=3$, we need $\left|\,\frac pq-\sqrt{223}\,\right|\lesssim\frac1{10q^2}$ and that can only happen when the next continuant is $13$ or $28$. This only happens on the over estimates; that is, when $\frac pq\gt\sqrt{223}$.

For example,
$(14)=14$ is an underestimate, but is not close enough: $223\cdot1^2-14^2=27$
$(14;1)=15$ is close enough, but it is an overestimate: $223\cdot1^2-15^2=-2$
$(14,1,13)=\frac{209}{14}$ is an underestimate, but is not close enough: $223\cdot14^2-209^2=27$
$(14;1,13,1)=\frac{224}{15}$ is close enough, but it is an over estimate: $223\cdot15^2-224^2=-1$


I discuss the explicit mapping to ideals at http://math.blogoverflow.com/2014/08/23/binary-quadratic-forms-over-the-rational-integers-and-class-numbers-of-quadratic-%EF%AC%81elds/

There will not be a proof of this just by congruences. The trouble is that $-3$ really is represented by a form in the principal genus of this discriminant. The whole genus (these are Gauss-Legendre reduced forms, $ax^2 + bxy + c y^2$ with $ac <0$ and $b > |a+c|$) is

1.             1          28         -27   cycle length             4
2.            -3          28           9   cycle length             6
3.             9          28          -3   cycle length             6

the other genus is

4.            -1          28          27   cycle length             4
5.             3          28          -9   cycle length             6
6.            -9          28           3   cycle length             6

The formalism of quadratic fields does not distinguish between the two genera. This calculation of classes came out very well, in that Dirichlet's description of Gauss composition, for binary quadratic forms, shows that $$ \langle A,B,A^2 \rangle $$ or $Ax^2 + B xy+ A^2 y^2,$ with $\gcd(A,B) =1,$ satisfies group laws $$ \langle A,B,A^2 \rangle^2 = \langle A^2,B,A \rangle $$ $$ \langle A,B,A^2 \rangle^3 = \langle A^3,B,1 \rangle = \langle 1,B,A^3 \rangle $$ and here $A=-3, B=28.$

Continued fractions, for $\sqrt {223},$ show the small numbers (absolute value up to $\sqrt {223}$ ) that are primitively represented by $x^2 - 223 y^2,$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell 223

0  form   1 28 -27   delta  -1
1  form   -27 26 2   delta  13
2  form   2 26 -27   delta  -1
3  form   -27 28 1   delta  28
4  form   1 28 -27

and that list is just $1,2,$


I have an interesting story on this equation.

The Minkowski bound of $\mathbb{Z}[\sqrt{223}]$ is less than $15$. So the ideal class group is generated by the prime ideals above $3$ and $11$. Since $16^2-223=11\cdot3$, this means the class group is generated by the prime ideal above $3$. That means, your problem is equivalent to ask whether the class number of $\mathbb{Z}[\sqrt{223}]$ is $1$ or not.

Now the big point comes in, the cubic polynomial $x^3-5x^2+2$ has discriminant $4\cdot223$. Then $\mathbb{Q}(\sqrt{223},\alpha)/\mathbb{Q}(\sqrt{223})$ is an everywhere unramified extension, where $\alpha$ is a root of this polynomial. So the class number of $\mathbb{Q}(\sqrt{223})$ is divides by $3$ by class field theory. (In fact, it is $3$).

The polynomial $x^3-5x^2+2$ was found by total accident.


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