Find integer points on this equation

[littlemathquark]

Homework Statement

Find integer points on $x^2-xy-6x+y^2+5y+6=0$

Relevant Equations

Find integer points on $x^2-xy-6x+y^2+5y+6=0$

One of my solution is:
$x^2-xy-6x+y^2+5y+6=(x-y)^2+(x-6)^2+(y+5)^2=49$ and by trying integer $(x,y)$ points are $(0,-2),(0,-3),(3,1),(3,-3),(4,-2),(4,1)$

Another idea $y^2+(5-x)y+x^2-6x+6=0$ or $x^2-(y+6)x+y^2+5y+6=0$ and solving quadratic equation considering to integer solutions.

My question is: Are there other solutions that I can't see and is there a criterion to (number of ) find integer-coordinate points on the curve defined by the given equation?

 

[Mark44]

One of my solution is: $x^2-xy-6x+y^2+5y+6=(x-y)^2+(x-6)^2+(y+5)^2=49=0$ and by trying integer $(x,y)$ points are $(0,-2),(0,-3),(3,1),(3,-3),(4,-2),(4,1)$

You can't get from $x^2 - xy \dots$ to $(x - y)^2 \dots$.
Also, the part $\dots = 49 = 0$ is almost certainly a typo in which you hit = instead of +.

 

[fresh_42]

Ignoring the fact that you lost a $xy$ term in your "equation", let's consult WA:
https://www.wolframalpha.com/input?i=x^2-xy-6x+y^2+5y+6=0

It not only lists the integer solutions, it also shows how to prove that there aren't others by inspecting the root expressions.

 

[littlemathquark]

Ignoring the fact that you lost a $xy$ term in your "equation", let's consult WA:
https://www.wolframalpha.com/input?i=x^2-xy-6x+y^2+5y+6=0

It not only lists the integer solutions, it also shows how to prove that there aren't others by inspecting the root expressions.

But why the solution $(4,1)$ is not in the solution list?

 

[littlemathquark]

You can't get from $x^2 - xy \dots$ to $(x - y)^2 \dots$.
Also, the part $\dots = 49 = 0$ is almost certainly a typo in which you hit = instead of +.

$2(x^2-xy-6x+y^2+5y+6)=0*2$
$(x-y)^2+x^2-12x+y^2+10y+12=0$

 

[fresh_42]

But why the solution $(4,1)$ is not in the solution list?

I guess they only display five solutions. It is not difficult to complete the calculations:
\begin{align*}
a^2&=-3x^2+14x+1\, , \,2y=\pm a+x-5\\
3x&=7\pm \sqrt{52-3a^2}\\
3x&\in 7\pm \{2,5,7\}=\{0,2,5,9,12,14\}\\
(x,a)&\in \{(0,\pm 1),(3,\pm 4),(4,\pm 3)\} \\
(x,y)&\in \{(0,-2),(0,-3),(3,1),(3,-3),(4,-2),(4,1)\}
\end{align*}
It needs a little bit of effort from your side.

And if you click on "More Solutions" you will find all six solutions.

 

[littlemathquark]

I also considered using Pick's theorem, but I abandoned the idea because I didn't know the number of lattice/integer points inside the ellipse. Actually, I'm looking for a more general theory that gives the number of lattice points on a curve.

 

[fresh_42]

I also considered using Pick's theorem, but I abandoned the idea because I didn't know the number of lattice/integer points inside the ellipse. Actually, I'm looking for a more general theory that gives the number of lattice points on a curve.

Here are some papers I found by searching "lattice points on a curve + pdf"
https://math24.wordpress.com/wp-content/uploads/2013/02/lattice-point.pdf
https://www.worldscientific.com/doi/abs/10.1142/S1793042124500994
https://core.ac.uk/download/pdf/82591791.pdf

I doubt that there is one-solution-fits-all.

 

[littlemathquark]

Another solution. I used "Fermat sum of two squares theorem". What about solution?

The equation can be written as $(x-y)^2+(x-6)^2+(y+5)^2=49$ by multiplying main equation by 2. From this equation, it is clear that $y\le1$.
Theorem (Fermat's theorem on sums of two squares): A prime number p > 2 is the sum of two squares if and only if p ≡ 1 (mod 4).
For $y = 1$, the right side of the equation ( I mean $(x-1)^2+(x-6)^2=49-36=13$) is p = 13 = 4k + 1, so the equation has solutions. By trial and error, these solutions are found to be x = 3 and x = 4.

Theorem (theorem on sums of two squares): For a composite number containing a prime factor of the form 4k + 3 to be expressible as the sum of two squares, the exponents of these prime factors must be even. Only composite numbers containing prime factors of the form 4k + 1 can be written as the sum of two squares.

Now, let's continue using this theorem:

For $y = 0$, the right side of the equation is $24 = 2^3 · 3$, and since the exponent of the factor 3, which is of the form 4k + 3, is not even, it cannot be written as the sum of two squares. So, the equation has no solution.

For $y = -1$, the right side is 33 = 3 · 11, and for the same reason, 33 cannot be written as the sum of two squares.

For $y = -2$, the right side is $40 = 2^3 · 5$, and since 5 is of the form 4k + 1, 40 can be written as the sum of two squares. By trial and error, x = 0 and x = 34 are found.

For $y = -3$, the right side is $45 = 2^3 · 5$, and for the same reason, 45 can be written as the sum of two squares. By trial and error, x = 0 and x = 3 are found.

For $y = -4$, the equation $(x+4)^2+(x-6)^2=48$ does not represent a real circle.

Accordingly, all solutions are $(3, 1), (4, 1), (0, -2), (4, -2), (0, -3), (3, -3)$.