20211116, 08:49  #1 
Jun 2021
110011_{2} Posts 
A small question from the lost book
I'm found this in the old book, that imply a solution; there are missing pages
Let How to find the smallest value of epsilon for given p? Or x and y for the given p and epsilon say less then sqrt(p)??? Last fiddled with by RomanM on 20211116 at 09:25 Reason: typo ans>and 
20211116, 14:18  #2 
"Robert Gerbicz"
Oct 2005
Hungary
1575_{10} Posts 
For any p the e values is not lower bounded:
set say x=1, then x^2*(x^3+y^3)=y^3+1 goes to inf for y>inf. For your 2nd question: about solving the equation for a given n=p+e: x^2*(x^3+y^3)=n first method: factorize n, then try each squared divisor as for d=x^2 you can solve it: y=(n/dx^3)^(1/3) ofcourse check if y is an integer or not (you had two choices for x: x=+sqrt(d)). 2nd method: economical solution, find only all p<10^9 that is a prime divisor of n (and then the exact primepower p^e divisor), after this again try the squared divisors, the idea of this approach is that for random(!) n values it is unlikely that there is even a single p value for that p^2n (you have less than 1e9 probability for this). 
20211116, 15:05  #3 
Jun 2021
63_{8} Posts 
Thank You for the answer! The above arose from this
but m^5+m^2 is narrow number to match any chosen p. If m=x/y; all much more interesting; if we took numerators of fractions that turned out by this substitution, the one (==1) turns to be square. Last fiddled with by RomanM on 20211116 at 15:06 
20211118, 11:11  #4 
Jun 2021
3·17 Posts 
i.e. A^2==B^2 mod p, p=x^2(x^3+y^3); May be this too obviosly and known??? Let y=1 and we can easy factor m^5+1 for any m. 
20211118, 13:17  #5 
Feb 2017
Nowhere
2^{5}·11·17 Posts 

20211118, 15:42  #6  
Jun 2021
3×17 Posts 
Quote:
nor m^5+1, m^2*(m^3+1)= m^5+m^2 I make an error here if we take some big m and compute mod((2*m^3+1)^2, m^5+m^2)=1, than compute mod((2*m^3+1)^2, m^5+m^2+1)=A, A will be relative small compare to m^5+m^2+1, for p~10^270, A~10^54 compare this to QS 10^134 so instead of sieve we can build such n+eps=p, where p number to factor, eps  small number, n  number for those we can build the left part as above i.e. B^2==1 mod n Last fiddled with by RomanM on 20211118 at 16:11 Reason: *** 

20211118, 16:16  #7 
Feb 2017
Nowhere
2^{5}·11·17 Posts 
And you'll see that the congruence you asked about holds as a polynomial congruence (mod x^{2}(x^{3} + y^{3})).

20211118, 17:43  #8 
Jun 2021
3×17 Posts 
I'm slowly came to understand this! Thanks!
this originlal formula. mod((a*m^3+1/2*a*C)^2,m^5+C*m^21/4*a^2*C^24/a^2*m)=A a=2, C=1 and we have a silly me with super obvios (2m^3+1)^2=blah blah)) whatever, main idea still the same instead of sieve we can build alot of such n+eps=p, where p number to factor, eps  small number, 1 is the best)), n  number of some special form, for those we can build the left part as above i.e. B^2==1 mod n, then compute our beloved little residual, factor them and do the same math as in QS Last fiddled with by RomanM on 20211118 at 18:10 
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