Quadratic Equation with 2 Variables

Question

I have this Equation :
$xy + y^2 = 19 + 2x^2$

I need to find every $(x , y)$ where $x$ and $y$ are integers.

I first tried to solve it by dividing everything by $y^2$ , but $19$ creates a problem , because we will get term $19/y^2$ , which is cumbersome & may or may not be Integer.

I next tried to solve for $x$ and check when is square root of Discriminant integer, but not sure how to check that.

Can anyone help me out ?

Can you post what you've tried so far as part of the question? It will help people to give you more useful answers. — Chris Lewis, Sep 18 '23 at 11:21
@NikoMolecule you said that one approach "creates a problem" and another you're "not sure" about. What problem? What aren't you sure about? — Chris Lewis, Sep 18 '23 at 11:24
This is the equation of a conic section. You can reduce it to standard form by a change of variables. — Gribouillis, Sep 18 '23 at 11:24
Hint: you want to use the fact $x$ and $y$ are integers. Both your approaches ignore this - you're looking at solving for reals. Can you see a way to rearrange and factorise the equation? — Chris Lewis, Sep 18 '23 at 11:25
@ChrisLewis i factorised equation to this:
$$(y + \sqrt{19})(y - \sqrt{19}) = x(2x - y)$$ — NikoMolecule, Sep 18 '23 at 11:31
@NikoMolecule See the answer by Prem for the factorisation. In your comment, again, you avoided the fact that $x$ and $y$ are integers. The point is, you know everything there is to know about integer factors of $19$, so the fact you're after integer solutions is more "important" than the fact it's a quadratic equation. — Chris Lewis, Sep 18 '23 at 13:02

Prem · Accepted Answer · 2023-09-18T11:54:11.237

2

This is Diophantine Equation , involving Quadratic terms.

Let us take the given Equation :
$xy+y^2=19+2x^2\tag{1}$

Move the terms around , to get :
$xy+y^2-2x^2=19$
$xy-x^2+y^2-x^2=19$

We can now factorize this :
$x(y-x)+(y+x)(y-x)=19$
$[x+y+x](y-x)=19\tag{2}$

We have 2 Integers , which give Prime Number $19$ when multiplied.
Hence , the Numbers must be either $\pm 1$ or $\pm 19$.
$[x+y+x]=+19 \land (y-x)=+1\tag{3}$
$[x+y+x]=+1 \land (y-x)=+19\tag{4}$
$[x+y+x]=-19 \land (y-x)=-1\tag{5}$
$[x+y+x]=-1 \land (y-x)=-19\tag{6}$

Solve these ~~two~~ four Sets of Equations to get the Solutions , selecting Integer Solutions.

With (3) & (4) , We will get Solutions $(6,7)$ & $(-6,13)$ which are Integers.

With (5) & (6) , User TonyK gives the Solutions $(-6,-7)$ & $(6,-13)$ which are Integers too.

edited Sep 18 '23 at 11:54

answered Sep 18 '23 at 11:30

Prem

9,669

2

Thank you so much, that was very helpfull <3 – NikoMolecule Sep 18 '23 at 11:32
1

You mean, the numbers must be $\pm 1$ or $\pm 19$. This gives the additional solutions $(-6,-7)$ and $(6,-13)$. – TonyK Sep 18 '23 at 11:44
Oh yes , I missed those $\pm$ Cases , @TonyK , thank you ! I will add those Cases now ! – Prem Sep 18 '23 at 11:49

Quadratic Equation with 2 Variables

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