How to find all pairs $(a,b)$ of natural numbers that satisfy $a^2-2b^2=-73696$ and $\gcd(a,b)=4$
The only thing I was able to do is to change the equation to
$a_1^2-2b_1^2=-4606$ with $\gcd(a_1,b_1)=1$ This gives that $2|a_1$ so the equation becomes $2a_2^2-b_1^2=-2303$ ,$2a_2=a_1$
I can't find useful bound or factorization to complete , Can someone help me please ?
On
Use the famous identity
$\begin{array}\\ (u^2-nv^2)(x^2-ny^2) &=u^2x^2-nu^2y^2-nv^2x^2+n^2v^2y^2\\ &=u^2x^2+n^2v^2y^2-n(u^2y^2+v^2x^2)\\ &=u^2x^2+2nu^2v^2x^2y^2+n^2v^2y^2-n(u^2y^2+2u^2v^2x^2y^2+v^2x^2)\\ &=(ux+nvy)^2-n(uy+vx)^2\\ \end{array} $
Therefore, if there is a solution to $x^2-ny^2=1$, then, if $u^2-nv^2=d$, there are an infinite number of solutions to $u_m^2-nv_m^2=d$ starting with $u_1=u, v_1=v$ and $u_{m+1}=u_mx+nv_my, v_{m+1}=u_my+v_mx $.
On
This problem lives naturally in $\mathbb Z[\sqrt2]$ with norm $N(x+y\sqrt2)=x^2-2y^2$ and Brahmagupta's identity is very useful; it says that the norm is multiplicative.
Write $a^2-2b^2=-73696 = 2^4 \cdot 7^2 \cdot (-2) \cdot 47$.
If $x_1^2-2y_1^2=-2$, $x_2^2-2y_2^2=47$, then we can get $x_3^2-2y_3^2=(-2) \cdot 47$ using Brahmagupta's identity. Finally we get $a=2^2\cdot 7 \cdot x_4$ and $b=2^2\cdot 7 \cdot y_4$.
Carrying out this plan, we start with $(x_1,y_1)=(0,1)$ and $(x_2,y_2)=(7,1)$, found by inspection. Then $(x_3,y_3)=(2,7)$ and $(a,b)=(56,196)$.
You get infinitely many solutions of $a^2-2b^2=-73696$ by using Brahmagupta's identity with even powers of $1+\sqrt2$, that is, powers of $3+2\sqrt2$.
Since $\gcd(56,196)=4$, this continues to hold because $(x+y\sqrt2)(3+2\sqrt2)$ can be expressed as $$ \begin{pmatrix} 3 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ and the matrix has determinant $1$.
There are infinitely many solutions and they come with a linear degree two recursion. I was finding $x^2 - 2 y^2 = 2303$
The first observation is a generator for the (oriented) automorphism group of the quadratic form $$ x^2 - 2 y^2 = (3x+4y)^2 - 2(2x+3y)^2 $$ As your target $2303$ is composite we need six families of solutions as $$ x_{n+6} = 3x_n + 4 y_n $$ $$ y_{n+6} = 2x_n + 3 y_n $$
These have parallel recurrences $$x_{n+12} = 6 x_{n+6} - x_n $$ $$y_{n+12} = 6 y_{n+6} - y_n $$
I think I will include the inequality part, this is how the business gets proved. The solutions that my program, below, calls "seeds" are those $x,y >0 $ such that either $3x-4y \leq 0$ or $-2x+3y \leq 0.$ The $3x-4y$ condition causes no restrictions, but the second, $y \leq \frac{2x}{3},$ gives firm upper bounds on both $x,y.$ Finding where the line intersects the hyperbola, we learn $x < \sqrt {20727} \approx 143.9687,$ with the $y$ bound two thirds of that. And there are exactly six such fundamental solutions.
let me type in the lists
as the $x$ begin $$ 49, 55, 65, 71, 89, 119, 175, 241, 319, 361, 479, 665, 1001, 1391 $$
while the $y$ begin
$$7, 19, 31, 37, 53, 77, 119, 167, 223, 253, 337, 469, 707, 983, 1307, 1481, 1969, 2737, 4123, 5731,$$
let me throw in the pairs list which has the line numbers I am calling $n$