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 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)))^2-1/6 ((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))) (4+10 (1+I Sqrt[3]) (2/(11+3 Sqrt[69]))^(1/3)+I 2^(2/3) (I+Sqrt[3]) (11+3 Sqrt[69])^(1/3))+1/144 (-4 I+10 (-I+Sqrt[3]) (2/(11+3 Sqrt[69]))^(1/3)+2^(2/3) (I+Sqrt[3]) (11+3 Sqrt[69])^(1/3))^2+(1/(-54+8 Sqrt[69]))(\[Sqrt](1/6 (14568 2^(2/3) (11+3 Sqrt[69])^(1/3)-14568 I 2^(2/3) Sqrt[3] (11+3 Sqrt[69])^(1/3)+5058 I 2^(2/3) Sqrt[23] (11+3 Sqrt[69])^(1/3)-1686 2^(2/3) Sqrt[69] (11+3 Sqrt[69])^(1/3)+18879 2^(1/3) (11+3 Sqrt[69])^(2/3)+18879 I 2^(1/3) Sqrt[3] (11+3 Sqrt[69])^(2/3)-6849 I 2^(1/3) Sqrt[23] (11+3 Sqrt[69])^(2/3)-2283 2^(1/3) Sqrt[69] (11+3 Sqrt[69])^(2/3)-4496 3^(5/6) Sqrt[23] (-27+4 Sqrt[69])^(1/3)-2224 3^(1/6) Sqrt[23] (-27+4 Sqrt[69])^(2/3)+38848 (-81+12 Sqrt[69])^(1/3)+5504 (-81+12 Sqrt[69])^(2/3)+(27-3 Sqrt[69])^(2/3) (2^(1/3) (12043+4283 I Sqrt[23])+4 (498-263 I Sqrt[23]) (11+3 Sqrt[69])^(1/3))-3^(1/6) (9-Sqrt[69])^(2/3) (2^(1/3) (36129 I+4283 Sqrt[23])+4 (-1494 I+263 Sqrt[23]) (11+3 Sqrt[69])^(1/3))+2 (27-3 Sqrt[69])^(1/3) (2^(2/3) (2044-339 I Sqrt[23])+(-5396-1926 I Sqrt[23]) (11+3 Sqrt[69])^(2/3))+2 3^(5/6) (9-Sqrt[69])^(1/3) (2^(2/3) (2044 I-113 Sqrt[23])+(5396 I+642 Sqrt[23]) (11+3 Sqrt[69])^(2/3))))))
to display on a page. (FullSimplify just spits back the sextic.)
Date: 2016-12-29 09:56 From: Warut Roonguthai <warut822@gmail.com>
For the first one, 4*(x^6-3*x+3) = (2*x^3-3*x+3)^2 + 3*(2*x^2-x-1)^2. Warut
So there are no real roots. But does this tell me what the roots actually are? (E.g., In[996].) --rwg And are there ≥822 GMail customers named Warut?
On Thu, Dec 29, 2016 at 5:55 AM, Bill Gosper <billgosper@gmail.com> wrote:
irreducible, indecomposable, nonreciprocal, solvable: In[996]:= Factor@Decompose[MinimalPolynomial[I/Sqrt[3]-(I (1/2 (9 I-2 Sqrt[3]+3 3^(1/4) Sqrt[-4 I-3 Sqrt[3]]))^(1/3))/3^(2/3)+(I+Sqrt[3])/(2^(2/3) (27 I-6 Sqrt[3]+9 3^(1/4) Sqrt[-4 I-3 Sqrt[3]])^(1/3)),x],x] Out[996]= {3-3 x+x^6}
In[997]:= Factor@Decompose[MinimalPolynomial[1/12 (2-2^(2/3) (-29+3 Sqrt[321])^(1/3)+2^(2/3) (29+3 Sqrt[321])^(1/3)-2^(5/6) Sqrt[3 (7 2^(1/3)+(83-3 Sqrt[321])^(1/3)+(83+3 Sqrt[321])^(1/3))]),x],x] Out[997]= {-1-x-x^5+x^6}