On Jun 26, 12:30 am, Jan Burse <janbu...@fastmail.fm> wrote: > Andrew Tomazos schrieb: > > > On Jun 22, 4:01 am, Jan Burse <janbu...@fastmail.fm> wrote: > >> b*b = (c+a) * (c-a) > >> n*q*m*n*q*m = n*q*n * q*m*m (factorization of b, c+a, c-a sic!) > > > In this step you have used: > > > b = n*q*m > > c+a = q*n*n > > c-a = q*m*m > > > How do you know that there exists n, q and m such that these three > > equalities hold? > > -Andrew. > > By some pingeonhole principle. > > n, q and m are classes of factors of b. If a class c is empty > then c=1. If a class c contains factors j, k, etc.. then c=j*k*... > So the n, q and m are not itself necessary prime, they group > certain prime number factors. > > Each factor of b occurs twice in b*b, so b*b=n*q*m*n*q*m is trivial. > > And thus (c+a)*(c-a)=n*q*m*n*q*m follows also, since factors are unique, > the same classes also occure in the product (c+a)*(c-a). > > 1) Now if a factor does not occur in (c+a) then it occurs twice in > (c-a), this is the class m. > > 2) Further if a factor does not occur in (c-a) then it occurs twice in > (c+a), this is the class n. > > 3) If a factor does not (not occur in (c+a) and not occur in (c-a)), > then it occurs in (c+a) and it occurs in (c-a), this is the class q.
We can assume that theorems about the nature of integer factorization have previously been established - but which ones specifically would we use to prove your above statements?
I think it would be beneficial if we could compute q, n and m directly.
Perhaps we can define a divisble_by function. So for example: