Solve $x^2+2xy-782y=0$ diophantine equation

Question

I'm trying to solve $x^2+2xy-782y=0$ diophantine equation.

With these steps:

a) $(*4)$; $4x^2+8xy-3128y=0$

b) $(+/-4y^2)$; $4x^2+8xy+4y^2-3128y-(4y^2)=0$

c) Reducing; $(2x+2y)^2+(-3128y)-(4y^2) = 0$

d) $u=2x+2y$; $u^2-4y^2-3128y=0$

e) $(*-4)$; $-4u^2+16y^2+12512y=0$

f) $(+/-1564^2)$; $-4u^2+16y^2+12512y+2446096-2446096 =0$

g) Reducing; $(-4)u^2+(4y+1564)^2-2446096=0$

h) $v=4y+1564$; $(-4)u^2+v^2-2446096=0$

And that's a Pell's equation:

$v^2-4u^2=2446096$

That I can't solve because '4' is a perfect square.

Are there other methods to solve the original equation?

Thanks

Answer 1

$$y=\dfrac{x^2}{2(391-x)}$$ which must be an integer

As the denominator is even, so will be the numerator

Let $x=2z$

$$\implies y=\dfrac{2z^2}{391-2z}=-z+\dfrac{391z}{391-2z}$$

If $d$ divides $391-2z,391z$

$d$ must divide $2(391z)+391(391-2z)=391^2$

Now $391=17\cdot23$

So the $391-2z$ must divide $391^2=(17\cdot23)^2$

and $391-2z$ must divide $391z$ as well

Can you take it from here?

Answer 2

Use $$v^2-4u^2=(v-2u)(v+2u)$$ and solve the equation systems $$v-2u=a$$ $$v+2u=b$$ for every pair $(a,b)$ with $ab=2446096$. To find those pairs, determine all divisors (also the negative ones) of $2446096$.

Note that the mentioned equation is NOT a Pell equation because $4$ is a perfect square.

Answer 3

First at all a little transformation: $x_1^2+2x_1y_1-2\cdot17\cdot23\space y_1=0\Rightarrow (x_1,y_1)=(2x,2y)$ and we solve $$x^2+2xy-17\cdot23\space y=0\iff y=\frac{x^2}{17\cdot23-2x}; x\in \mathbb Z$$ It follows $$x(x+2y)\equiv 0\pmod {17\cdot23}$$ Hence the possibilities $$\begin{cases}x=17m\space\space\text {or}\space\space x+2y=17m\\x=23n\space\space\text{or}\space\space x+2y=23n\\x=391m\space\space\text {or}\space \space x+2y=391m \end{cases}$$ This gives right away three infinite sets of rational solutions by the choice of $x=17m$, $x=23n$ and $x=391m$ (we don’t pay attention to the other three possibilities with the factor $(x+2y)$.

One has $$x=17m\Rightarrow y=\frac{17m^2}{23-2m}\qquad (1)$$ $$x=23m\Rightarrow y=\frac{23m^2}{17-2m}\qquad (2)$$ $$x=391m\Rightarrow y=\frac{391m^2}{1-2m}\qquad (3)$$ However we need integer solutions; look at the easier equalities $(3)$: it follows the equation

►For equality $(3)$$$\frac{391m^2}{1-2m}=n; \space(m,n)\in \mathbb Z^2$$ which have solutions for small values of $m$.

We may use again congruences, so one has $n(1-2m)\equiv 0\pmod{17\cdot23}$. We get, for example, $$(m,n)\in\{(1,-391),(-8,1472),(9,-1863,(-11,2057)…..\}$$ ►For equality $(1)$, $$\frac{17m^2}{23-2m}=n$$ we get, for example, $$(m,n)\in\{(3,9),(11,2057),(20,-400), …..\}$$ ►For equality $(2)$, we have, for example, $$(m,n)\in\{(-3,9),(8,1472),(9,-1863),(17,391), …..\}$$ I could not find a general parameterization as I wanted. This seems very hard or impossible to achieve.