Intereting Posts

Zeros of Fourier transform of a function in $C$
The unique loop $L$ (quasigroup with unit) of order $5$ satisfying $x^2 = 1$ for all $x \in L$
Summability vs Unconditional Convergence
Complex number: calculate $(1 + i)^n$.
Do zeros of uniformly convergent function sequences also converge?
What is the logic/rationale behind the vector cross product?
Showing $f'(x) = f(x)$ implies an exponential function
The equation $x^n+y^n+z^n=u^n+v^n+w^n=p$, where $p$ is prime.
Variety generated by finite fields
Is the $\mathfrak m$-adic completion of a radical ideal again a radical ideal?
$K(u,v)$ is a simple extension of fields if $u$ is separable
Open mathematical questions for which we really, really have no idea what the answer is
Sets and classes
A square integrable martingale has orthogonal increments
Prove projection is self adjoint if and only if kernel and image are orthogonal complements

Given distinct primes $p$ and $q$, both congruent to $1 \pmod 4$, such that $$\left(\frac{p}{q}\right) = 1$$ and obviously also $$\left(\frac{q}{p}\right) = 1$$ is it possible for the fundamental unit of $\mathcal{O}_{\mathbb{Q}(\sqrt{pq})}$ to have a norm of $1$?

There is Theorem eleven point five point seven in saban alaca williams something strange going on with my keyboard in this paragraph< sorry>

- Is the algebraic closure of a $p$-adic field complete
- Quotient of the ring of integers of a quadratic field by the ideal generated by a split integer prime.
- What are the integers $n$ such that $\mathbb{Z}$ is integrally closed?
- Proof of Stickelberger’s Theorem
- Counting the Number of Integral Solutions to $x^2+dy^2 = n$
- The discriminant of an integral binary quadratic form and the discriminant of a quadratic number field

- How to calculate the ideal class group of a quadratic number field?
- A condition for an odd prime to be represented by a binary quadratic form of a given discriminant
- Quotient of the ring of integers of a quadratic field by the ideal generated by a split integer prime.
- On Diophantine equations of the form $x^4-n^2y^4=z^2$
- Integral solutions of hyperboloid $x^2+y^2-z^2=1$
- Convert from Nested Square Roots to Sum of Square Roots
- Equivalent Definitions of Ideal Norm
- Euclidean domain $\mathbb{Z}$
- Why is $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$?
- Are all algebraic integers with absolute value 1 roots of unity?

Added: for small numbers, it seems about one out of three numbers $n=pq,$ twelve out of the smallest forty four, with primes $p \equiv q \equiv 1 \pmod 4$ and $(p|q)= (q|p) = 1,$ do give integer solutions to $x^2 – n y^2 = -1.$ The first few are

$$ 145 = 5 \cdot 29, \; \; x=12, y=1 $$

$$ 445 = 5 \cdot 89, \; \; x=4662, y=221 $$

$$ 901 = 17 \cdot 53, \; \; x=30, y=1 $$

$$ 1145 = 5 \cdot 229, \; \; x=1252, y=37 $$

$$ 1313 = 13 \cdot 101, \; \; x=616, y=17 $$

$$ 1745 = 5 \cdot 349, \; \; x=4052, y=97 $$

$$ 2249 = 13 \cdot 173, \; \; x=197140, y=4157 $$

It seems to hold steady at about one out of three, there were $5820$ successes out of the first $18000$ such numbers. There were $99284$ out of the first $300000.$

**HOWEVER:**

$205 = 5 \cdot 41,$ and there are no integer solutions to

$$ x^2 – 205 y^2 = -1. $$ More to the point, there are no integer solutions to

$$ x^2 + x y – 51 y^2 = -1. $$

With reference to the second screen capture below, with $x=20$ and $y=3,$

$$ 20^2 + 20 \cdot 3 – 51 \cdot 3^2 = 1. $$

Or to

$$ x^2 + 13 x y – 9 y^2 = -1. $$

$221 = 13 \cdot 17,$ and there are no integer solutions to

$$ x^2 – 221 y^2 = -1. $$ There are also no integer solutions to

$$ x^2 + x y – 55 y^2 = -1. $$

With reference to the third screen capture below, with $x=7$ and $y=1,$

$$ 7^2 + 7 \cdot 1 – 55 \cdot 1^2 = 1. $$

Or to

$$ x^2 + 13 x y – 13 y^2 = -1. $$

Screen captures from http://www.numbertheory.org/php/unit.html

Ummm. here is the 205 thing in my language:

```
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 1 -51
0 form 1 1 -51 delta 0
1 form -51 -1 1 delta 6
2 form 1 13 -9
-1 -6
0 -1
To Return
-1 6
0 -1
0 form 1 13 -9 delta -1 ambiguous
1 form -9 5 5 delta 1
2 form 5 5 -9 delta -1 ambiguous
3 form -9 13 1 delta 13
4 form 1 13 -9
form 1 x^2 + 13 x y -9 y^2
minimum was 1rep x = 1 y = 0 disc 205 dSqrt 14.317821063 M_Ratio 205
Automorph, written on right of Gram matrix:
2 27
3 41
=========================================
```

And here is 221

```
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 1 1 -55
0 form 1 1 -55 delta 0
1 form -55 -1 1 delta 6
2 form 1 13 -13
-1 -6
0 -1
To Return
-1 6
0 -1
0 form 1 13 -13 delta -1 ambiguous
1 form -13 13 1 delta 13 ambiguous
2 form 1 13 -13
form 1 x^2 + 13 x y -13 y^2
minimum was 1rep x = 1 y = 0 disc 221 dSqrt 14.866068747 M_Ratio 221
Automorph, written on right of Gram matrix:
-1 -13
-1 -14
=========================================
```

My Theorem 1 is Theorem 11.5.5 on page 279 of Alaca and Williams. My Theorem 2 is Theorem 11.5.7 on page 286 of Alaca and Williams. It turns out Theorem 2 is due to Dirichlet, 1834.

I keep forgetting why $x^2 + xy – k y^2$ dominates $u^2 – (4k+1)v^2.$ One step: take $x = u-v, y = 2v.$ Then

$$ x^2 + xy – k y^2 = u^2 – 2 uv + v^2 + 2 u v – 2 v^2 – 4 k v^2 = u^2 – (4k+1) v^2. $$

THEOREM 1: With prime $p \equiv 1 \pmod 4,$ there is always a solution to $$ x^2 – p y^2 = -1 $$ in integers. The proof is from Mordell, Diophantine Equations, pages 55-56.

PROOF: Take the smallest integer pair $T>1,U >0$ such that $$ T^2 – p U^2 = 1. $$

We know that $T$ is odd and $U$ is even. So, we have the integer equation

$$ \left( \frac{T+1}{2} \right) \left( \frac{T-1}{2} \right) = p \left( \frac{U}{2} \right)^2. $$

We have $$ \gcd \left( \left( \frac{T+1}{2} \right), \left( \frac{T-1}{2} \right) \right) = 1. $$

Indeed,

$$ \left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1. $$

There are now two cases, by unique factorization in integers:

$$ \mbox{(A):} \; \; \; \left( \frac{T+1}{2} \right) = p a^2, \; \; \left( \frac{T-1}{2} \right) = b^2 $$

$$ \mbox{(B):} \; \; \; \left( \frac{T+1}{2} \right) = a^2, \; \; \left( \frac{T-1}{2} \right) = p b^2 $$

Now, in case (B), we find that $(a,b)$ are smaller than $(T,U),$ but $T \geq 3, a > 1,$ and $a^2 – p b^2 = 1.$ This is a contradiction, as our hypothesis is that $(T,U)$ is minimal.

As a result, case (A) holds, with evident $$p a^2 – b^2 = \left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1, $$

so

$$ b^2 – p a^2 = -1. $$

THEOREM 2: With primes $p \neq q,$ with $p \equiv q \equiv 1 \pmod 4$ and Legendre $(p|q)=(q|p) = -1,$ there is always a solution to $$ x^2 – pq y^2 = -1 $$ in integers. The proof is from Mordell, Diophantine Equations, pages 55-56.

PROOF: Take the smallest integer pair $T>1,U >0$ such that $$ T^2 – pq U^2 = 1. $$

We know that $T$ is odd and $U$ is even. So, we have the integer equation

$$ \left( \frac{T+1}{2} \right) \left( \frac{T-1}{2} \right) = pq \left( \frac{U}{2} \right)^2. $$

We have $$ \gcd \left( \left( \frac{T+1}{2} \right), \left( \frac{T-1}{2} \right) \right) = 1. $$

There are now four cases, by unique factorization in integers:

$$ \mbox{(1):} \; \; \; \left( \frac{T+1}{2} \right) = a^2, \; \; \left( \frac{T-1}{2} \right) = pq b^2 $$

$$ \mbox{(2):} \; \; \; \left( \frac{T+1}{2} \right) = p a^2, \; \; \left( \frac{T-1}{2} \right) = q b^2 $$

$$ \mbox{(3):} \; \; \; \left( \frac{T+1}{2} \right) = q a^2, \; \; \left( \frac{T-1}{2} \right) = p b^2 $$

$$ \mbox{(4):} \; \; \; \left( \frac{T+1}{2} \right) = pq a^2, \; \; \left( \frac{T-1}{2} \right) = b^2 $$

Now, in case (1), we find that $(a,b)$ are smaller than $(T,U),$ but $T \geq 3, a > 1,$ and $a^2 – pq b^2 = 1.$ This is a contradiction, as our hypothesis is that $(T,U)$ is minimal.

In case $(2),$ we have

$$ p a^2 – q b^2 = 1. $$

$$ p a^2 \equiv 1 \pmod q, $$ so $a$ is nonzero mod $q,$ then

$$ p \equiv \left( \frac{1}{a} \right)^2 \pmod q. $$

This contradicts the hypothesis $(p|q) = -1.$

In case $(3),$ we have

$$ q a^2 – p b^2 = 1. $$

$$ q a^2 \equiv 1 \pmod p, $$ so $a$ is nonzero mod $p,$ then

$$ q \equiv \left( \frac{1}{a} \right)^2 \pmod p. $$

This contradicts the hypothesis $(q|p) = -1.$

As a result, case (4) holds, with evident $$pq a^2 – b^2 = \left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1, $$

so

$$ b^2 – pq a^2 = -1. $$

The viewpoint of real quadratic fields an norms of fundamental units is more concerned with $x^2 + xy – k y^2,$ where $4k+1 = pq$ in the second theorem. However, we showed above that the existence of a solution to $u^2 – (4k+1)v^2 = -1$ gives an immediate construction for a solution to $x^2 + xy – k y^2=-1,$ namely $x=u-v, y=2v.$

EXTRA

David Speyer reminded me of something Kaplansky wrote out for me, years and years ago. From David’s comments, I now understand what Kap was trying to show me. If $x^2 + x y – 2 k y^2 = -1,$ then $(2x+y)^2 – (8k+1)y^2 = -4. $ This is impossible $\pmod 8$ unless $y$ is even, in which case $(x + \frac{y}{2})^2 – (8k+1) \left( \frac{y}{2}\right)^2 = -1.$

There is more to it when we have $x^2 + x y – k y^2 = -1$ with odd $k.$ Take

$$ u = \frac{ 2 x^3 +3 x^2 y + (6k+3)x y^2 + (3k+1)y^3}{2}, $$

$$ v = \frac{3 x^2 y + 3 x y^2 + (k+1)y^3}{2}. $$

$$ u^2 – (4k+1) v^2 = -1, $$ since

$$ u^2 – (4k+1) v^2 = \left( x^2 + x y – k y^2 \right)^3. $$

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