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roots of third degree polynomial

asked 2019-06-09 09:56:26 -0500

santoshi gravatar image

roots of polynomial x^3+7x+25 over field F(37)

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What have you tried?

John Palmieri gravatar imageJohn Palmieri ( 2019-06-09 13:02:56 -0500 )edit

Homework ?

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2019-06-10 13:05:01 -0500 )edit

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answered 2019-06-13 08:46:30 -0500

dan_fulea gravatar image

It may be well a homework, but maybe one should give the solution, since part of it is knowing the right method to be used in the right context. The following initializes the polynomial ring $\Bbb F_{37}[X]$ in the transcendental variable $X$, an other one as the $x$ set by default, and maybe not the $x$ in the OP. Then having the polynomial over the right field we simply ask for its roots. A second solution would be to use a "polynomial expression" using the variable x, which exists by default (or create any other), then use the method roots also specifying as optional parameter the ring for the roots.

sage: var('x');
sage: R.<X> = PolynomialRing(GF(37))
sage: (x^3 + 7*x + 25).roots(ring=GF(37))
[(5, 1)]
sage: (x^3 + 7*x + 25).roots(ring=GF(37), multiplicities=False)
[5]
sage: (X^3 + 7*X + 25).roots()
[(5, 1)]
sage: (X^3 + 7*X + 25).roots(multiplicities=False)
[5]
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With a field of size 37, you can also use brute force: f = X^3 + 7*X + 25 and then [y for y in GF(37) if f(y) == 0].

John Palmieri gravatar imageJohn Palmieri ( 2019-06-13 12:55:37 -0500 )edit

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Asked: 2019-06-09 09:56:26 -0500

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Last updated: Jun 13