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-   -   A small question from the lost book (https://www.mersenneforum.org/showthread.php?t=27327)

RomanM 2021-11-16 08:49

A small question from the lost book
 
I'm found this in the old book, that imply a solution; there are missing pages

Let [TEX]x^{2}(x^{3}+y^{3})=p+\epsilon \quad x,y,p,\epsilon \in\ Z[/TEX]

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)???

R. Gerbicz 2021-11-16 14:18

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).

RomanM 2021-11-16 15:05

Thank You for the answer! The above arose from this [TEX](2m^{3}\pm1)^{2}\equiv1\; mod\;m^{5}\pm m^{2}[/TEX]
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.

RomanM 2021-11-18 11:11

[TEX](2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z[/TEX]

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.

Dr Sardonicus 2021-11-18 13:17

[QUOTE=RomanM;593353][TEX](2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z[/TEX]

i.e. A^2==B^2 mod p, p=x^2(x^3+y^3); May be this too obviosly and known???
<snip>[/QUOTE]Factor (2x[sup]3[/sup] + y[sup]3[/sup])[sup]2[/sup] - y[sup]6[/sup] as the difference of two squares.

RomanM 2021-11-18 15:42

[QUOTE=Dr Sardonicus;593365]Factor (2x[sup]3[/sup] + y[sup]3[/sup])[sup]2[/sup] - y[sup]6[/sup] as the difference of two squares.[/QUOTE]
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+[B]1[/B])=A, A will be relative small compare to m^5+m^2+[B]1[/B],
for p~10^270, A~10^54 compare this to QS 10^134
so instead of [B]sieve[/B] we can [B]build[/B] 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

Dr Sardonicus 2021-11-18 16:16

[QUOTE=RomanM;593379][QUOTE=RomanM;593353][TEX](2x^{3}+y^{3})^{2}\equiv y^{6}\; mod\;x^{2}(x^{3}+y^{3}); \; y\neq0,\; x,y\in Z[/TEX][/quote][QUOTE=Dr Sardonicus;593365]Factor (2x[sup]3[/sup] + y[sup]3[/sup])[sup]2[/sup] - y[sup]6[/sup] as the difference of two squares.[/QUOTE]And what?[/QUOTE]And you'll see that the congruence you asked about holds as a polynomial congruence (mod x[sup]2[/sup](x[sup]3[/sup] + y[sup]3[/sup])).

RomanM 2021-11-18 17:43

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


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