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Forum: Miscellaneous Math 2021-11-18, 17:43
Replies: 7
Views: 620
Posted By RomanM
I'm slowly came to understand this! Thanks! ...

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...
Forum: Miscellaneous Math 2021-11-18, 15:42
Replies: 7
Views: 620
Posted By RomanM
And what? nor m^5+1, m^2*(m^3+1)= m^5+m^2 I...

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...
Forum: Miscellaneous Math 2021-11-18, 11:11
Replies: 7
Views: 620
Posted By RomanM
(2x^{3}+y^{3})^{2}\equiv y^{6}\;...

(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...
Forum: Miscellaneous Math 2021-11-16, 15:05
Replies: 7
Views: 620
Posted By RomanM
Thank You for the answer! The above arose from...

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...
Forum: Miscellaneous Math 2021-11-16, 08:49
Replies: 7
Views: 620
Posted By RomanM
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...
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