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You are not logged in. #1 20121020 02:30:45
Factorising a^n + kb^nIs there a method of deducing whether or not a polynomial of the form is factorisable?For instance, we can see that and but something like is not factorisable. Is there a way to tell if we can factorise something of this form? Is there an easy way to do this, or would your best bet be just to write a general factorisation and solve for your general coefficients? (for instance, writing one factor as (x^2 + ax + b) and finding a and b or something. #2 20121020 05:07:32
Re: Factorising a^n + kb^nYou can try cyclotomic polynomials. I remember bobbym said they can be used for deriving such identities, so... The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20121020 05:24:08
Re: Factorising a^n + kb^nYes, I saw that too, but I have no idea how to use those here. I can't find anything that discusses multivariable cyclotomic polynomials. #4 20121020 07:55:01
Re: Factorising a^n + kb^nWell, Last edited by anonimnystefy (20121020 08:04:11) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #6 20121020 08:03:57
Re: Factorising a^n + kb^nFixed it. You can treat a/b like only one variable, and then calculate ((a/b)+k)^n and then multiply by b^n. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20121020 08:12:43
Re: Factorising a^n + kb^nAre you sure that is correct? I tested it with SophieGermain's identity (k = 4, n = 4) and I am not getting a^4 + 4b^4... unless I went wrong somewhere. #8 20121020 10:49:24
Re: Factorising a^n + kb^nHi all; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20121020 21:22:49
Re: Factorising a^n + kb^nShouldn't it be #10 20121020 22:48:08
Re: Factorising a^n + kb^nThat is correct  but I'm more interested in integer factors. In other words, the above is not desirable as a does not always divide b (for a = 2, b = 3 for instance). For example, suppose that I wanted to deduce whether or not something was prime  finding a factorisation with fractions in it might not help. #11 20121021 07:27:44
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #12 20121021 07:38:21
Re: Factorising a^n + kb^nWhy, though? I can't see why, for example, if a²  b² factorises to (ab)(a+b), that one condition is that b divides a. Yet that is also of the form a^{n} + kb^{n}. #13 20121021 07:40:22
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #15 20121021 07:45:40
Re: Factorising a^n + kb^nDo you still have that computer program you used to compute factorisations? #16 20121021 07:47:00
Re: Factorising a^n + kb^nHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #17 20121021 07:51:18
Re: Factorising a^n + kb^nSo you were able to pull that a^6 + 8b^6 factorisation off the top of your head? #18 20121021 07:54:55
Re: Factorising a^n + kb^nHi; To start, did you read post #2? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #19 20121021 08:00:50
Re: Factorising a^n + kb^nYes but you never told me how you used cyclotomic polynomials to do that... #20 20121021 08:04:55
Re: Factorising a^n + kb^nMore than a century ago, I came across a book with a big chart of Aurifeuillian Factorizations. I was amazed. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #21 20121021 08:17:39
Re: Factorising a^n + kb^nHow does that relate to this? The wiki article is saying it is a factorisation of the form 2^{4n+2} + 1. #22 20121021 08:20:15
Re: Factorising a^n + kb^nThe point is I had tables of them like a table of integrals or sums. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #24 20121021 08:24:32
Re: Factorising a^n + kb^nSo, I am guessing this book might be a bit difficult for me to find... #25 20121021 08:26:51
Re: Factorising a^n + kb^nYes, I am ancient. You know that old quote:
In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 