Find out, with no resolution, whether or not the second degree diophantine equations with 2 unknowns have solutions

Question

I want to find out the existence of the solutions in diophantine equations of the style:

$$-259y ^2+ 2400yx + 1817y + 2122x = $$ $$1057364602723981500371957207036553770637547302056514367123547565680640946707606178926389130616$$

The point is that solving it with current methods (elliptic curves) takes a long time, so I want to find out whether or not it has solutions.

What I have already tried:

1. If the GCD of the coefficients does not divide the independent term, then it has no solutions. Even if I divided it, it may be that the equation has solutions or it may not have them

2. Shaping the $x$ and the $y$: for example $x,y$ pairs and come to a contradiction (sometimes it works, sometimes it doesn't)

My questions are:

-Do you know of any procedure that allows deciding whether or not it has solutions given any second-degree diophantine equation with two unknowns?

-The equation above is of a hyperbolic type, is it possible to modify it to make it an elliptical type? If yes, how do I do it?

Answer 1

$$-259y ^2+ 2400yx + 1817y + 2122x = $$ $$1057364602723981500371957207036553770637547302056514367123547565680640946707606178926389130616$$

$$\implies \Bigl(1440000 x + 1364999\Bigr)^2-\Bigl(600(2400 x - 518 y + 1817)\Bigr)^2=$$

$$394354702231936140378725159936353094296979641774997598362398300096251847484068800492386090829229590001$$

For solving need factorize RHS.

Any quadratic diophantine equations with 2 unknowns can simplify to 3 type:

1. Pell equation like $x^2-dy^2=\pm c$

2. difference of squares like $x^2-y^2=c$

3. "quadratic Thue" equation like $x^2+dy^2=c$

where $d,c\in\mathbb{N}$.

Answer 2

The general form of the "type" of the equation is (4) in

https://mathworld.wolfram.com/PellEquation.html

And there are algorithms to reduce the equation. The idea is to homogenize, thus getting an equation of the shape $$ q(x,y,z):= a_{11}x^2 +a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{23}yz+2a_{13}xz = 0\ . $$ See also https://encyclopediaofmath.org/wiki/Quadratic_form.

Solutions $[x:y:z]$ with $z\ne 0$ of $q=0$ (over $\Bbb Q$ or equivalently over $\Bbb Z$) correspond to solutions over $\Bbb Q$

The second question has a clear answer, the discriminant is an invariant of a ternary quadratic form and there is no linear transformation / no base change switching its sign. A "naive definition of type" is in this homogeneous version unclear, because after grouping squares in some linear combinations $X,Y,Z$ of $x,y,z$ we can write equivalently for $a,b,c>0$ $$ \begin{aligned} aX^2 + bY^2 &= cZ^2\ ,\\ aX^2 &= cZ^2-bY^2\ , \end{aligned} $$ and dehomogenizing w.r.t. $Z$ on the one side, w.r.t. $X$ on the other side, we get different "naive types".

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Note that the given diophantine equation is not related to elliptic curves. (There the world is much complicated. There is no local-to-global (Hasse) principle holding. An elliptic curve may have solutions in all place / localizations, without having solutions over $\Bbb Q$. In this given quadratic case the Hasse principle is a theorem.)

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To solve it, let us denote by $N$ the big number on the R.H.S. of the given equation, so we can write it in one line: $$ 259y^2 - 2400xy - 2122x - 1817y + N =0\ . $$ We multiply with $4\cdot 259$ and first group squares in all terms involving $y$, getting successively: $$ \begin{aligned} 0 &= 259y^2 - 2400xy - 2122x - 1817y + N \ ,\\ 0 &= 4\cdot259^2y^2 - 2\cdot 259\cdot 4800xy - 4\cdot 2122\cdot 259x - 4\cdot259\cdot 1817y + 1036N \ ,\\ 0 &= (518y - 2400x - 1817)^2 \\ &\qquad -5760000x^2 - 10919992x + 1036N - 1817^2 \ ,\\ 0 &= (518y - 2400x - 1817)^2 \\ &\qquad -\left(2400^2x^2 + 2\cdot \underbrace{\frac {10919992}{2\cdot 2400 }}_{:=s} \cdot 2400x +s^2\right) \\ &\qquad\qquad + s^2 + 1036N - 1817^2 \ ,\\ \end{aligned} $$ and because of $s=1364999/600$, we multiply with the denominator $600^2$ of $s^2$, thus getting: $$ \tag{$\dagger$} \\ 0= 600^2(518y - 2400x - 1817)^2 - (600\cdot 2400 x + 1817^2)^2 + \underbrace{ 1364999^2 +600^2( 1036N - 3301489)}_{:=M} \ . $$ The "(conceptually) simple problem" of solving a quadratic diofantine equation turns out to lead to the factorization of a "big number", $$ \tiny M= 394354702231936140378725159936353094296979641774997598362398300096251847484068800492386090829229590001\ , $$ so that we can match factors in the equivalent equation $(\dagger\dagger)$: $$ \tag{$\dagger\dagger$} \\ 259(1200y + 1061)(2880000x - 310800y + 2455199) =M\ . $$ And $M$ is indeed divisible by $259$, so let $M'=M/259$. $$ \tiny M' = 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139\ . $$ It turns out that we are here in a lucky situation. Writing $M'=(\pm1)\cdot(\pm M')$ and trying to match in all four cases the two factors to the factorization of the L.H.S. above leads to solutions in $\Bbb Q$, and one pairing leaves over $\Bbb Z$. So we do not need the factorization for the existence question. (Finding all solution would require the prime factors. This would be practically a more complicated question. Yes, here we could use elliptic curves.)

Optically we see that $M'+1061$ has last digits $00$, so this must be the chance we have. (All other pairings fail when taken modulo $100$.)

From here on, i will use sage to compute and type the steps, and find one solution of the initial problem over $\Bbb Z$.

sage: N = 1057364602723981500371957207036553770637547302056514367123547565680640946707606178926389130616
sage: var('x,y');
sage: def f(x,y): return -259*y^2 + 2400*y*x + 1817*y + 2122*x - N
sage: R.<X,Y> = PolynomialRing(ZZ)
sage: M = 1364999^2 + 600^2*( 1036*N - 3301489 )
sage: def g(x,y): return 600^2*(518*Y - 2400*X -1817)^2 - (600*2400*X + 1364999)^2 + M
sage: 4*259*600^2*f(X,Y) + g(X,Y)
0
sage: factor(g(X,Y) - M)
7 * 37 * (1200*Y + 1061) * (-2880000*X + 310800*Y - 2455199)

sage: MM = ZZ(M/259)                                                                                                          
sage: MM                                                                                                                      
1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139

sage: eq1 = 1200*y + 1061 == -MM
sage: eq2 = -2880000*x + 310800*y - 2455199 == 1
sage: sol = solve([eq1, eq2], [x,y], solution_dict=True)[0]
sage: sol
{x: -136928716052755604298168458311233713297562375616318610542499409755643002598635000170967392649039,
 y: -1268837523268777800446348648443864524765056762467817240548257078816769136049127414711666958910006}
sage: x0, y0 = sol[x], sol[y]
sage: f(x0, y0)
0

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The integer solution is explicitly: $$ \tiny \begin{aligned} x &= -136928716052755604298168458311233713297562375616318610542499409755643002598635000170967392649039 \ ,\\ y &= -1268837523268777800446348648443864524765056762467817240548257078816769136049127414711666958910006 \ . \end{aligned} $$

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Later EDIT:

Let us verify the above condition:

x0, y0 = sol[x], sol[y]
print(f"x0 = {x0}")
print(f"y0 = {y0}")
print(f"f(x0, y0) is {f(x0, y0)}")

Result:

x0 = -136928716052755604298168458311233713297562375616318610542499409755643002598635000170967392649039
y0 = -1268837523268777800446348648443864524765056762467817240548257078816769136049127414711666958910006
f(x0, y0) is 0

So these values verify the given equation.

This verification is here, since there is a comment below claiming that the result of inserting the above solution in the LHS of the given equation is not the RHS, but an other one, one of the shape $10573\dots8340$. OK, let us check the last digit. Then for the above solution working modulo ten we have $$ -259y_0 ^2+ 2400y_0x_0 + 1817y_0 + 2122x_0 \equiv 1\cdot 4^2 + 0 + 7\cdot 4+2\cdot 1=6+8+2\equiv 6\ .$$ So the last digit is $6$, not zero. (Please check the own computation, it has to be done using a proper high precision calculator like pari/gp.)

$$