they make a little more sense factoring a term I was leaving expanded…http://imgur.com/a/nDuao

]]>Dear all

RSA numbers linked to the pythagorean triples associated with Erdös-Straus decompositions…. N = 2099999959 is factorised in less than 10 / 100th of a second almost “instantly” The general characterization is given by :

– p = s*N*r/((2*a*s – r*N)*( s+r+t))

– q = ((2*a*s – r*N)*( s+r+t))/(s*r)

and/or

– p = s*N*r/((2*a*s – r*N)*( s-r+t))

– q = ((2*a*s – r*N)*( s-r+t))/(s*r)

Best regards

I believe I can show that the upper bound on in the Type I solution case can be improved to , unless , in which . I’m not sure, maybe this is what the authors had in mind when it was said that the bounds could be improved slightly. Anyway, for what it’s worth, here is my argument. I use the notation from the paper, Lemma 2.7.

First, observe that , since and . Next, observe that . This follows by starting from (2.3) in the paper, , and, since , it follows that . Next, observe that . This follows from , by substituting . Next, observe that . This follows from , by substituting . Thus the solution .

Assume now that . Then , say where . So .

But since , . If , then since , , a contradiction. So assume . Then , so . But , so this means . This means . But since , this means that , and, since , it must be that , so = 1 or 2. If , then , which is clearly impossible. If , then and , so , and . Since , this case can only happen when .

]]>In our paper we have some lower bounds on the average number of solutions in this case (which we call the k=4 case), but we were unable to match them with good upper bounds. There are of course more solutions in the k=4 case than the k=3 case (and in particular, there is obviously at least one solution for each n) but by the same token, they are harder to parameterise efficiently.

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