Solving quadratic equations over polynomials

AARRRGH

Hello

I hope someone can help me with this.

I have the following equation:

y^2 + xy = x^3 + a*x^2 + b

I want to solve it. I know what a and b are (3 and 5) but I do not know what x and y are.

If I let f(x) = x^3 + a*x^2 + b (i.e. the RHS) I can convert my equation to:

y^2 + xy + f(x) = 0

What is the simpiest way to solve this equation?

Do I have to just choose random x values and then brute force every possible y value?

Any help appreciated.

Thanks.

Thomas_S_Hunterson

If you sub in a value for x, you get the corresponding value for y. If you sub in a value for y, you get the corresponding x value. There is no definite x/y as such.

To find the roots (which I think is what you're trying to do, although it's not technically a function), let y=0 and solve for x so basically solve x^3 + 3x^2 + 5 = 0, which has 1 real and 2 complex roots by the look of it.

MathsManiac

As indicated by Sean K, the equation you've given has infinitely many solutions: it represents a curve in the plane, and any point on the curve will satisfy the equation and is therefore a solution.

If you're interested in real-valued solutions only, then if you pick any value of x (provided it's bigger than about -3.6) you'll get two corresponding values of y, namely:
y = (-1/2)*x+(1/2)*sqrt(13*x^2+4*x^3+20) and
y = (-1/2)*x-(1/2)*sqrt(13*x^2+4*x^3+20).

Attached is an image of part of the graph.
dublindude_graph.jpg

I don't believe there are any integer solutions, so you'd have great fun finding a solution by trial and error!

LeixlipRed

This is obviously some sort of Elliptic Curve related to cryptography thing judging by your previous posts. Are you trying to find the points on the curve over a finite field or something like that?

Fremen

"DublinDude" becomes "AAAARGH".

Thesis stressing you out, dublindude?

thebaldsoprano

Hi AAARGH,

Not exactly sure on the context or what you mean by 'solve', but if you're referring to a polynomial ring, a theorem along these lines might help:

http://mathworld.wolfram.com/EisensteinsIrreducibilityCriterion.html

Getting that nasty bastard of an equation into a 'nice' poly is a little too interesting for me right now though and making me a little glad not to be in college any more

AARRRGH

Thanks for the replies everyone.

I implemented a horrible brute force solution for the moment. I will return to trying to figure this out when my thesis is done.

Fremen wrote: »

"DublinDude" becomes "AAAARGH".

Thesis stressing you out, dublindude?

The worst is over thank God!

I managed to successfully implement a polynomial version of elliptic curve cryptography. My code is unbelievably inefficient, but it works...

Thanks for the help everyone!