solving maths elliptic equation
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Hello everyone,
I need to solve an equation in matlab as such:
y2=x3+aX+b Given
4a3+27b2≠0
Thanks in advance, regards, Anupam
4 件のコメント
Walter Roberson
2013 年 12 月 26 日
That does not appear to be an elliptic to me.
John D'Errico
2013 年 12 月 26 日
編集済み: John D'Errico
2013 年 12 月 26 日
To me, it seems closer to cryptic than elliptic. Perhaps a rubric, or a cube from Rubic? Maybe even aseptic or dyspeptic. But in the end? Just a cubic. Now I must go, no longer a goldbrick.
Image Analyst
2013 年 12 月 26 日
Which of the 7 variables (y2, x3, X, a, a3, b, and b2) do you have values for, and which do you not have values for and need to solve for? What's elliptic? Everything looks linear to me, unless you didn't put in proper symbols for multiplication and exponentiation (* and ^), but I guess at first I should assume you know what you're doing/saying.
Walter Roberson
2013 年 12 月 26 日
It appears to me that the graph has to do with conic sections. Conic sections are not always ellipses.
回答 (5 件)
Anupam
2013 年 12 月 27 日
0 投票
1 件のコメント
Walter Roberson
2013 年 12 月 27 日
Is it perhaps over integers instead of over reals ? Because in reals it is
y = +/- sqrt(x^3+a*x+b)
and the constraint doesn't matter to that.
Roger Stafford
2013 年 12 月 27 日
編集済み: Roger Stafford
2013 年 12 月 27 日
As John has already pointed out, this is a "cubic" equation in the unknown x, not an "elliptic" equation. As he has also pointed out, it can be solved numerically using matlab's 'roots' function.
If you are interested in an explicit formula in terms of general values for a and b, you should read up on the theory of cubic equations. It was first discovered by Scipione Del Ferro some time in the fifteen hundreds. The following web sites, as well as many text books, explain that theory:
http://en.wikipedia.org/wiki/Cubic_function
http://mathworld.wolfram.com/CubicFormula.html
You can search Google for other presentations of the theory.
The condition that you mention, 4*a^3+27*b^2 ~= 0, excludes the case where all three roots are real but two or all three are multiple roots. Can you possibly tell us how such a condition relates to the generation of keys for cryptographic purposes?
0 件のコメント
Anupam
2013 年 12 月 28 日
0 投票
1 件のコメント
Walter Roberson
2013 年 12 月 28 日
Sorry, for legal reasons, we cannot discuss Elliptic Curve Encryption on Mathworks' servers.
Roger Stafford
2013 年 12 月 28 日
編集済み: Roger Stafford
2013 年 12 月 28 日
In my opinion, where you stated "x^3+ax+b=0" you misled us into thinking you wished to solve that cubic equation, and the "y^2" part remained a mystery. Now I understand that you are indeed dealing with elliptic curves of the type y^2 = x^3+a*x+b, but you still have not made it clear what you mean by the statement "solve the above equation for the pair of keys". What exactly do you mean by that? There is an infinite continuum of points (x,y) that lie on such a curve.
I have been looking at the web site
http://en.wikipedia.org/wiki/Elliptic_curve
which speaks of an operation in which two points, P and Q, on such a curve uniquely determine a third distinct point R which must also lie on the curve and at the same time must lie on the straight line through P and Q. Is that actually the problem you are posing here? If so, you should make that clear. We shouldn't have to make guesses as to what you are asking!
Note: The web site
http://mathworld.wolfram.com/EllipticCurve.html
has a solution for this problem with equations (10) and (11).
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