hyperbolic paraboloid w/ roots at x!=0, y = (8-x^2)/x

[madmath123]

I have a hyperbolic paraboloid with the formula
x^2+x y-8 with roots at x!=0, y = (8-x^2)/xhow do you find all integer roots?
I know that four of the roots are at x= 1, -1, 8, -8 and that there are four other integer roots. is there a method to this or do i just have to brute force it?

 

[DrPhil]

I have a hyperbolic paraboloid with the formula
x^2+x y-8 with roots at x!=0, y = (8-x^2)/x how do you find all integer roots?
I know that four of the roots are at x= 1, -1, 8, -8 and that there are four other integer roots. is there a method to this or do i just have to brute force it?

I don't know what an "integer root" is. The only roots I see are \(\displaystyle \pm\sqrt{8}\). If you would show your work I might understand what you are trying to do.

Are you looking perhaps for integer values of y=k? In that case (8-x^2) = kx, or

\(\displaystyle x^2 + kx - 8 = 0\)

For roots of that quadratic to rational, the discriminant \(\displaystyle k^2+32\) must be a perfect square; e.g., k = ±2, x=1±6.

My first inclination would have been to get rid of the inclination - that is, rotate the coordinate system to eliminate the cross term.

 

[madmath123]

Sorry I meant how do you find the integer solutions (cases were both x and y are integers) when values of x are –c =< x =< c. c is 8 in this particular problem. Note that any values of x > c or x < -c will yield a non-integer y value when plugged into y = (8-x^2)/x.

To find the solutions you first solve x^2+x y-8 for y. when you do that for this particular equation you get x!=0, y = (8-x^2)/x.Now this is where I am stuck. I noticed that when you change the c value then if x= +-c and x=+-1 y will be an integer value every time. These are not given by the rational discriminant. I also tested all values –c < x < c (-8 < x < 8) for this problem and found that when x = -4, 4, 2, -2, I get integer values for y as well.

Now the thing is if I had a large number c then this method will be really inefficient. So is there a way to find all integer solutions for the equation y = (8-x^2)/x. Note that y will not be an integer when x != -8, -4, -2, -1, 1, 2, 4, 8.If you are interested the y values that I got from plugging these x values in are y = 7, 2, -2, -7, 7, 2, -2, -7. Note that when x is matched with the corresponding y then x2+kx−8=0 is true just make k=y.

You said that I can find the integer values for y=k using the rational discriminant “Are you looking perhaps for integer values of y=k? In that case (8-x^2) = kx, or x2+kx−8=0 For roots of that quadratic to rational, the discriminant k2+32 must be a perfect square; e.g., k = ±2, x=1±6.”

Also while we are on the subject I see no method to finding where the discriminant is a perfect square except by using brute force (graphing calculator or something of the sort). Also the rational discriminant doesn’t give me the integer solutions.

 

[daon2]

Also while we are on the subject I see no method to finding where the discriminant is a perfect square except by using brute force (graphing calculator or something of the sort). Also the rational discriminant doesn’t give me the integer solutions.

\(\displaystyle k^2+32 = k^2(1+32/k^2)\). This cannot be a perfect square for \(\displaystyle |k|> \sqrt{32/3}\)

 

[madmath123]

\(\displaystyle k^2+32 = k^2(1+32/k^2)\). This cannot be a perfect square for \(\displaystyle |k|> \sqrt{32/3}\)

This is not true because there is a perfect square at 7 and 7 >[FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]32[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]3[/FONT]

 

[daon2]

This is not true because there is a perfect square at 7 and 7 >[FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]32[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]3[/FONT]

You are correct, it was my blunder, I must have mistakenly assumed 1+32/k^2 was an integer.

How about this: notice for \(\displaystyle k>15\), \(\displaystyle (k+1)^2 > k^2+32\), and no perfect squares can lie strictly between \(\displaystyle k^2\) and \(\displaystyle (k+1)^2\). But \(\displaystyle k^2<k^2+32 < (k+1)^2\)

 

[daon2]

Another solution

Let \(\displaystyle n^2=k^2+32\)

Then \(\displaystyle (n-k)(n+k)=n^2-k^2=32\)

For every factorization \(\displaystyle ab=32\), we have a system of Diophantine equations

\(\displaystyle n-k=a\)
\(\displaystyle n+k=b=32/a\)

So:

\(\displaystyle 2n = a+32/a \iff n=\dfrac{a^2+32}{2a}\)

Letting \(\displaystyle a\) vary through all factors of 32 (both plus and minus) and throwing out non-integral solutions you'll have a very short list to check.