[math-fun] Two little sextics,
Whoa. Can PARI (or GAP) reliably denest radicals? http://community.wolfram.com/groups/-/m/t/988309 Maple, Mathematica, and (presumably) Maxima cannot. --rwg Date: 2016-12-30 02:19 From: Warut Roonguthai <warut822@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com> The Galois group of x^6-3*x+3 told me that it had a quadratic subfield (but no cubic subfield). The form of the exact value of x told me which quadratic field. (Otherwise, some more guessing would be needed, or I had to solve it the hard way.) I computed the Galois group and factored over Q(sqrt(-3)) with PARI. Warut On Fri, Dec 30, 2016 at 3:57 PM, Bill Gosper <billgosper@gmail.com> wrote:
Date: 2016-12-29 22:20 From: Warut Roonguthai <warut822@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com>
4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2 tells that x^6-3*x+3 can be factored into two cubic over Q(sqrt(-3)).
BTW, 822 is my lucky number but doesn't seem to work very well so far. Hahaha Warut
How could it not be at least doubly informative?
DAWK! I plead senility. Ok, how did you find those squares? I see how to go the other way--i.e. from a factorization over √n to f = p² - n q². E.g., if f = the duodecic for x2 in http://mathworld.wolfram.com/SquareDissection.html , guessing Factor[f,Extension->√{-1,2,3}] immediately cracked it with n = 2. Then 4 - 32 x2 + 80 x2^2 - 32 x2^3 - 168 x2^4 + 288 x2^5 - 164 x2^6 - 16 x2^7 + 60 x2^8 - 16 x2^9 - 4 x2^10 + x2^12 == -32 (-1 + x2)^4 (-1 + 2 x2)^2 + (6 - 24 x2 + 28 x2^2 - 8 x2^3 - 2 x2^4 + x2^6)^2
Is this what you did? --rwg Sadly, Noam Elkies tells me the Q(√2) sextic factors don't solve in radicals.
On Fri, Dec 30, 2016 at 11:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
Two extremes: In[466]:= Factor@Decompose[%462,x] Out[466]= {-4+24 x+12 x^2+4 x^3-6 x^4+x^6}
In[467]:= PolySolve6@@%//FullSimplify
Out[467]= (*All 6 solutions*) {{x->(-2)^(1/3)-Sqrt[2]},{x->(-2)^(1/3)+Sqrt[2]},{x->-2^(1/3) ((-1)^(2/3)+2^(1/6))}, {x->Sqrt[2]+Root[2+#1^3&,2]},{x->-2^(1/3) (1+2^(1/6))},{x->2^(1/3) (-1+2^(1/6))}}
Whereas Simplify can barely reduce just this one solution (a mere unit !-) Out[436]= x->Root[1+3 #1+4 #1^2+2 #1^3+#1^6&,6] In[469]:= Simplify@%435[[6,1]] Out[469]= x->1/2 (1/3+(1-I Sqrt[3])/(2^(2/3) (27-3 Sqrt[69])^(1/3))+((1+I Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))+1/6 I (I+Sqrt[3]) (1/2(11+3 Sqrt[69]))^(1/3)+(5 (1+I Sqrt[3]))/(3 2^(2/3) (11+3 Sqrt[69])^(1/3))-((1-I [... chop ...] to display on a page. (FullSimplify just spits back the sextic.) [... chop ...]
I guess not. Happy New Year 2017 to all math-fun subscribers. Warut On Sat, Dec 31, 2016 at 5:37 AM, Bill Gosper <billgosper@gmail.com> wrote:
Whoa. Can PARI (or GAP) reliably denest radicals? http://community.wolfram.com/groups/-/m/t/988309 Maple, Mathematica, and (presumably) Maxima cannot. --rwg
Date: 2016-12-30 02:19 From: Warut Roonguthai <warut822@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com>
The Galois group of x^6-3*x+3 told me that it had a quadratic subfield (but no cubic subfield). The form of the exact value of x told me which quadratic field. (Otherwise, some more guessing would be needed, or I had to solve it the hard way.) I computed the Galois group and factored over Q(sqrt(-3)) with PARI.
Warut
On Fri, Dec 30, 2016 at 3:57 PM, Bill Gosper <billgosper@gmail.com> wrote:
Then 4 - 32 x2 + 80 x2^2 - 32 x2^3 - 168 x2^4 + 288 x2^5 - 164 x2^6 - 16 x2^7 + 60 x2^8 - 16 x2^9 - 4 x2^10 + x2^12 == -32 (-1 + x2)^4 (-1 + 2 x2)^2 + (6 - 24 x2 + 28 x2^2 - 8 x2^3 - 2 x2^4 + x2^6)^2
Is this what you did? --rwg Sadly, Noam Elkies tells me the Q(√2) sextic factors don't solve in radicals.
On Fri, Dec 30, 2016 at 11:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
Two extremes: In[466]:= Factor@Decompose[%462,x] Out[466]= {-4+24 x+12 x^2+4 x^3-6 x^4+x^6}
In[467]:= PolySolve6@@%//FullSimplify
Out[467]= (*All 6 solutions*) {{x->(-2)^(1/3)-Sqrt[2]},{x->(-2)^(1/3)+Sqrt[2]},{x->-2^(1/3) ((-1)^(2/3)+2^(1/6))}, {x->Sqrt[2]+Root[2+#1^3&,2]},{x->-2^(1/3) (1+2^(1/6))},{x->2^(1/3) (-1+2^(1/6))}}
Whereas Simplify can barely reduce just this one solution (a mere unit !-) Out[436]= x->Root[1+3 #1+4 #1^2+2 #1^3+#1^6&,6] In[469]:= Simplify@%435[[6,1]] Out[469]= x->1/2 (1/3+(1-I Sqrt[3])/(2^(2/3) (27-3 Sqrt[69])^(1/3))+((1+I Sqrt[3]) (1/2 (9-Sqrt[69]))^(1/3))/(2 3^(2/3))+1/6 I (I+Sqrt[3]) (1/2(11+3 Sqrt[69]))^(1/3)+(5 (1+I Sqrt[3]))/(3 2^(2/3) (11+3 Sqrt[69])^(1/3))-((1-I [... chop ...] to display on a page. (FullSimplify just spits back the sextic.) [... chop ...]
Date: 2016-12-29 22:20 From: Warut Roonguthai <warut822@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com>
4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2 tells that x^6-3*x+3 can be factored into two cubic over Q(sqrt(-3)).
BTW, 822 is my lucky number but doesn't seem to work very well so far. Hahaha Warut
How could it not be at least doubly informative?
DAWK! I plead senility. Ok, how did you find those squares? I see how to go the other way--i.e. from a factorization over √n to f = p² - n q². E.g., if f = the duodecic for x2 in http://mathworld.wolfram.com/SquareDissection.html , guessing Factor[f,Extension->√{-1,2,3}] immediately cracked it with n =
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