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.
Since the cases 1, 2 and 3 are exhaustive (*), therefore (c+a)=q*n*n and (c-a)=q*m*m.
(*) This is usually a proof figure called "proof by cases",
A -> B ~A -> B ------------------- B
For 1), 2) and 3) it is applied multiple times, plus some more stuff.