diophantine equationselementary-number-theorynumber theorypell-type-equations
For example, solve this : $x^2 +2x +3 = 10y^2+11y+12$ with x,y integers
I can reduce this equation into this : $Y^2-40X^2=-279$ with $X=x+1$ and $Y=20y+11$
Now I need to find the primitive solutions of this equation but i don't know how to do it. And these primitive solutions that make it possible to find general solutions must be in a very specific form. Here, $Y=11 $ mod $ 20$. Any help would be appreciated.
EDIT : I did a mistake in the basic equation. I changed it.
Some rearranging and reduction: \begin{align} 8x + 10y &\equiv 8 \pmod 7 \iff \\ x + 3y + 7x + 7y &\equiv 1 + 7 \pmod 7 \iff \\ x + 3y &\equiv 1 \pmod 7 \end{align}
The following will apply the algorithms for linear congruence equations and for linear Diophantine equations (because it was asked) to get solutions. I use the algorithms here because they are guaranteed to give the number of solutions and all solutions.
Solution as linear congruence equation
We can interpret this as linear congruence $a X \equiv b \pmod m$ in the integer unknown $X$: \begin{align} x + 3y &\equiv 1 \pmod 7 \iff \\ 7x + 3y &\equiv 1 + 6x \pmod 7 \iff \\ 3y & \equiv 1 + 6x \pmod 7 \quad (*) \end{align}
For this there exists an solution algorithm (link) to find all solutions from $\mathbb{Z}_m$.
The linear congruence $(*)$ has $d = \gcd(a,m) = \gcd(3, 7) = 1$ and $d \vert b$ and $d$ solutions (per fixed $x$) in $\mathbb{Z}_m$. One uses the extended Euclidean algorithm to find $s, t$ with $1 = 3s + 7t$, e.g. $s = -2$, $t = 1$. We get the only solution from $\mathbb{Z}_7$: $y = (-2) (1+6x) \bmod 7$ or: $$ y = (5 + 2x) \bmod 7 $$ Regarding $\mathbb{Z}$, the general solution is $$ (x,y) = (x, ((5 + 2x) \bmod 7) + 7k) \quad (k \in \mathbb{Z}) \quad (\#) $$
Some example solutions
The colour encodes different $k$ parameters.
$$ (x, y) \in \{ \ldots (-2, 1), (-1, 3), (0, 5), (1, 0), (2, 2), (3, 4), (4, 6), (5, 1), (6, 3), (7, 5), (8, 0), \ldots \} $$
Test
\begin{align} x + 3 y &\equiv 1 \pmod 7 \iff \\ x + 3(5 + 2x) &\equiv 1 \pmod 7 \iff \\ 7 x + 15 &\equiv 1 \pmod 7 \iff \\ 7 x + 14 &\equiv 0 \pmod 7 \iff \\ 0 &\equiv 0 \pmod 7 \end{align}
Solution as Diophantine equation
We interpret the congruence as linear Diophantine equation $a X + b Y = c$ in the integer unknowns $X$ and $Y$: $$ x + 3y \equiv 1 \pmod 7 \iff \\ x + 3y = 1 + 7 k \quad (k \in \mathbb{Z}) $$ For this there exists an algorithm as well (link).
We have $d = \gcd(a, b) = \gcd(1,3) = 1$. So $d \vert c = 1+7k$. We set $m = 1 + 7k$. Then we need $s, t$ for $s + 3t = 1$, e.g. $s=1$, $t = 0$. This gives the particular solution $x_0 = ms = 1 + 7k$, $y_0 = mt = 0$, thus $$ (x_0, y_0) = (1+7k, 0) $$ All solutions (for fixed $k \in \mathbb{Z}$) are $$ (x_n, y_n) = (x_0 + bn, y_0 - an) = (1+7k+3n,-n) \quad (n \in \mathbb{Z}) $$ E.g. for $k=0, n=1$ we have the solution $(4,-1)$. But we want to have a similar solution like $(\#)$, in terms of $x$:
Now $7k + 3n = x - 1$ is another Diophantine equation, because $d=\gcd(7,3) = 1$ and $d\vert x - 1$ we have infinite solutions. We get $s = 1$, $t=-2$ and $k_0 = x-1$, $n_0=-2(x-1)=2-2x$, $k_i = x-1+3i$, $n_i = 2-2x - 7i$.
So we have the solutions \begin{align} (x,y) &= (1+7k+3n,-n) \\ &= (1+7(x-1+3i)+3(2-2x-7i), -(2-2x-7i)) \\ &= (1+7x-7+21i+6-6x-21i, -2+2x+7i) \\ &= (x, 2x-2+7i) \quad (i \in \mathbb{Z}) \quad (\#\#) \end{align} So this looks more similar to $(\#)$. And $(5+2x)\bmod 7 = (2x + 5) \bmod 7 = (2x-2) \bmod 7$. Thus because of $2x-2 = 7q + ((2x-2) \bmod 7)$ for some $q$ we got the same solution sets, once we add multiples of $7$.
The solution looks the same, the colour encodes different $i$ parameters. So we see the parametrization is different, but set seems the same, looking at the $[0,6]\times[0,6]$ with its $7$ solutions.
Hint:
Write $x=2^p$ (why ?) then you get $2^{12p}\equiv 2^y\pmod{19}$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12p\equiv y\pmod{18}$ (why ?)
Best Answer
This is not a complete answer,
but I did some calculations and found the following solutions to $Y^2-40X^2=-279: $
$(Y,X)=$ $ (9,3), (19,4), (41,7), (61, 10), (119, 19), $ and $(189,30)$.
If $(Y,X)$ is a solution, then so are $(-Y,X), (Y,-X), $ $(-Y,-X),$
and $(19Y+120X,3Y+19X)$. [Note: $19^2-40\cdot3^2=1$.]
Then we need to take solutions where $Y=20y+11$,
such as $(-9,\pm3), (-189,\pm30), $ and $(531,\pm84)$, which yield solutions $(y,x)=$
$(-1,-4), (-1, 2), (-10,-31), (-10, 29), (26, -85), $ and $(26,83)$ to your original equation.