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February 26Edit

Prove that any positive integer, can be decomposed to at most norm-equal 2 elements in gauss integerEdit

Hello, I'm trying to prove that any positive integer, can be decomposed to at most norm-equal 2 elements in gauss integer. My thought, write down the number as integer product, and then decompose each prime. My problem that I'm unable to handle gaussian prime numbers such 3. --Exx8 (talk) 22:48, 26 February 2021 (UTC)

I'm confused. First, I assume you mean decompose as a product, in other words factor; you can also decompose things as a sum, so you need to specify. Also, every integer is also a Gaussian integer, so if X is an integer then X taken as the product of 1 factor is a decomposition of X. One is less than two so that would be a decomposition of X into at most two Gaussian integer factors. So did you mean exactly two factors? If so then it can't be done, there is no such factorization for 3. This is a consequence of unique factorization which holds in the Gaussian integers. --RDBury (talk) 02:34, 27 February 2021 (UTC)
Yes as a product. Decomposition to 1 number is not decomposition to 2 numbers.

I meant that if X=YZ, and YZ are equal in their norm, then there is no A,B such X=AB, |A|=|B|,A≠Y, B≠Z up to multiplication by unity.--Exx8 (talk) 06:52, 27 February 2021 (UTC)

I take the claim to be that if  , where the two factors of either product have equal norms, then the factorizations on either side of the equality can be transformed into each other by multiplying these factors by units. Now   fits the antecedent of the implication, yet there is no unit   such that  . The claim holds in the complex numbers:   has norm  . But it is not a Gaussian integer, so it is not a unit in the ring of Gaussian integers.  --Lambiam 08:31, 27 February 2021 (UTC)

February 27Edit

A proof that there is no q such: q²=k²+1 by gaussian integersEdit

I'm trying to prove that there is no a real integer q such: q²=k²+1 via gaussian integers. I know there is a simpler proof with just decomposing it to equation of 1. But does anyone see a gaussian integer related-proof? Thanks --Exx8 (talk) 22:50, 27 February 2021 (UTC)

The question is unclear. The equation   has a solution in   when   or   Do you mean, prove that the equation   has no general solution in   for all values of  ? Then it is sufficient to give just a single counterexample, like the unsolvability of   Or does the use of "real integer" in the question mean that we should understand the variables   and   to range over  , and is your question whether the (simple) proof that this has no non-trivial solutions in   can be made complicated by involving the Gaussian integers? I have no idea what "equation of 1" means, and what the "it" is that can be decomposed to this equation. If you want us to consider your questions, you should really put more effort in formulating them clearly.  --Lambiam 07:46, 28 February 2021 (UTC)

let  . prove that :   with gauss integer properties. meaning use gauss-integer properties and theorems.--Exx8 (talk) 08:10, 28 February 2021 (UTC)

The k=0 case is still an exception so you'd have to work around that. If you add k≠0 as an assumption then it's hard to see how Gaussian integers wouldn't be making a mountain out of a molehill. If q2-k2=1 then (q-k)(q+k)=1, q-k=q+k=±1, whence q=±1, k=0. --RDBury (talk) 10:15, 28 February 2021 (UTC)

Yes of course, it is for k>=1. I know the proof for with equations. Can you see how to solve with gaussian integers?--Exx8 (talk) 12:38, 28 February 2021 (UTC)

Expounding on RDB's proof, solutions in the integers are also solution in the Gaussian integers, so the absence of solutions in the latter ring implies absence of solutions in the integers. Starting from   we see that   and   are both units and each other's conjugates, which, next to the solution   given already, only leaves   The excursion through the complex plane did not lead over a formidable mountain, but is an unhelpful detour nevertheless.  --Lambiam 12:47, 28 February 2021 (UTC)
Is there any connection to that in the complex number, we have a decomposition into to linear polynomials? I wonder if it can be somehow related to the gaussian integers...--Exx8 (talk) 13:02, 28 February 2021 (UTC)
I don't see a connection. In the ring   the polynomial   can be factored as   while the same polynomial is irreducible in    --Lambiam 15:07, 28 February 2021 (UTC)
Well is there something unique with   is there a reason which says that we cannot decompose   somehow else?--Exx8 (talk) 16:29, 28 February 2021 (UTC)
Suppose   By equating the coefficients, we have that   and   so   and therefore   This means that   This is nothing special for this polynomial: in general, the factorization of a polynomial in   into a multiset of linear factors is unique up to multiplication by a scalar (polynomial of degree 0).  --Lambiam 18:23, 28 February 2021 (UTC)

March 2Edit

Notation questionEdit

All the following notation is gleaned from WP. Given a differentiable manifold M with tangent bundle TM and cotangent bundle TM, the set of sections of TM (also called vector fields) is denoted Γ(TM), and similarly of TM is denoted Γ(TM). The exterior algebra on Γ(TM) is denoted Ω(M) (the differential forms on M). Is there a suitable equivalent notation for the exterior algebra of Γ(TM), i.e., the dual of Ω(M)? —Quondum 13:45, 2 March 2021 (UTC)

March 5Edit