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 2021-11-16, 08:49 #1 RomanM   Jun 2021 3×17 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 $x^{2}(x^{3}+y^{3})=p+\epsilon \quad x,y,p,\epsilon \in\ Z$ 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 2021-11-16 at 09:25 Reason: typo ans->and
 2021-11-16, 14:18 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 62716 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/d-x^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^2|n (you have less than 1e-9 probability for this).
 2021-11-16, 15:05 #3 RomanM   Jun 2021 3×17 Posts Thank You for the answer! The above arose from this $(2m^{3}\pm1)^{2}\equiv1\; mod\;m^{5}\pm m^{2}$ 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 2021-11-16 at 15:06
 2021-11-18, 11:11 #4 RomanM   Jun 2021 3·17 Posts $(2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z$ 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.
2021-11-18, 13:17   #5
Dr Sardonicus

Feb 2017
Nowhere

175B16 Posts

Quote:
 Originally Posted by RomanM $(2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z$ i.e. A^2==B^2 mod p, p=x^2(x^3+y^3); May be this too obviosly and known???
Factor (2x3 + y3)2 - y6 as the difference of two squares.

2021-11-18, 15:42   #6
RomanM

Jun 2021

3·17 Posts

Quote:
 Originally Posted by Dr Sardonicus Factor (2x3 + y3)2 - y6 as the difference of two squares.
And what?
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 2021-11-18 at 16:11 Reason: ***

2021-11-18, 16:16   #7
Dr Sardonicus

Feb 2017
Nowhere

3·1,993 Posts

Quote:
Originally Posted by RomanM
Quote:
 Originally Posted by RomanM $(2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z$
Quote:
 Originally Posted by Dr Sardonicus Factor (2x3 + y3)2 - y6 as the difference of two squares.
And what?
And you'll see that the congruence you asked about holds as a polynomial congruence (mod x2(x3 + y3)).

 2021-11-18, 17:43 #8 RomanM   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^2-1/4*a^2*C^2-4/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 2021-11-18 at 18:10

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