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)[reply]

The question is unclear. The equation ${\displaystyle q^{2}=k^{2}+1}$ has a solution in ${\displaystyle q}$ when ${\displaystyle k=0}$ or ${\displaystyle k=i.}$ Do you mean, prove that the equation ${\displaystyle q^{2}=k^{2}+1}$ has no general solution in ${\displaystyle q}$ for all values of ${\displaystyle k}$? Then it is sufficient to give just a single counterexample, like the unsolvability of ${\displaystyle q^{2}=2.}$ Or does the use of "real integer" in the question mean that we should understand the variables ${\displaystyle q}$ and ${\displaystyle k}$ to range over ${\displaystyle \mathbb {Z} }$, and is your question whether the (simple) proof that this has no non-trivial solutions in ${\displaystyle \mathbb {Z} }$ 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)[reply]

let ${\displaystyle k\in \mathbb {Z} }$. prove that : ${\displaystyle \nexists q\in \mathbb {Z} .q^{2}=k^{2}+1}$ with gauss integer properties. meaning use gauss-integer properties and theorems.--Exx8 (talk) 08:10, 28 February 2021 (UTC)[reply]

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)[reply]

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)[reply]

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 ${\displaystyle (q+k)(q-k)=1,}$ we see that ${\displaystyle q+k}$ and ${\displaystyle q-k}$ are both units and each other's conjugates, which, next to the solution ${\displaystyle q=\pm 1,k=0}$ given already, only leaves ${\displaystyle q=0,k=\pm i.}$ 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)[reply]

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)[reply]

I don't see a connection. In the ring ${\displaystyle \mathbb {C} ,}$ the polynomial ${\displaystyle X^{2}+2}$ can be factored as ${\displaystyle (X+i{\sqrt {2}})(X-i{\sqrt {2}}),}$ while the same polynomial is irreducible in ${\displaystyle \mathbb {Z} [i].}$  --Lambiam 15:07, 28 February 2021 (UTC)[reply]

Well is there something unique with ${\displaystyle k^{2}+1=(k+i)(k-i)}$ is there a reason which says that we cannot decompose ${\displaystyle k^{2}+1}$ somehow else?--Exx8 (talk) 16:29, 28 February 2021 (UTC)[reply]

Suppose ${\displaystyle k^{2}{+}1=(k{+}a)(k{+}b).}$ By equating the coefficients, we have that ${\displaystyle a{+}b=0}$ and ${\displaystyle ab=1,}$ so ${\displaystyle b=-a.}$ and therefore ${\displaystyle a^{2}=-1.}$ This means that ${\displaystyle a=\pm i,b=\mp i.}$ This is nothing special for this polynomial: in general, the factorization of a polynomial in ${\displaystyle \mathbb {C} }$ 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)[reply]