Forum: Miscellaneous Math
2022-01-31, 20:05
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Replies: 19
Views: 1,152
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Forum: Miscellaneous Math
2022-01-24, 19:55
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Replies: 17
Views: 1,043
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Forum: Miscellaneous Math
2022-01-24, 10:47
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Replies: 17
Views: 1,043
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Forum: Miscellaneous Math
2021-12-22, 14:11
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Replies: 2
Views: 351
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Forum: Miscellaneous Math
2021-12-22, 13:59
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Replies: 2
Views: 351
An extension of NFS
Hi.
Both QS and NFS stuck with famouse A^2==B^2 mod p.
If we look on the A^3==B^3 mod p that do the split also (30% vs 50% for quadratic)
and in general to A^n==B^n*m & B^m=y mod p?? (1)
as a...
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Forum: Homework Help
2021-12-15, 18:47
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Replies: 2
Views: 981
Casus of x^3+1
Do You know that
(2*(x^2+1)^3-7))^2\equiv3^4\ mod\ x^3+1
for all x in Z, abs(x)>=5???
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Forum: Miscellaneous Math
2021-12-01, 16:37
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Replies: 1
Views: 298
Approximation of r by m^(1/n)
r\pm\epsilon=\sqrt[n]{m}
where r - real root of polynomial P(r), order >=6, and m, n - integers, and P(r+/-eps)<1
P.S. I'm suspect that the simpler the question look like, the less likely it is to...
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Forum: Miscellaneous Math
2021-11-18, 17:43
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Replies: 7
Views: 624
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...
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Forum: Miscellaneous Math
2021-11-18, 15:42
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Replies: 7
Views: 624
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...
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Forum: Miscellaneous Math
2021-11-18, 11:11
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Replies: 7
Views: 624
(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...
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Forum: Miscellaneous Math
2021-11-16, 17:18
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Replies: 5
Views: 992
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Forum: Miscellaneous Math
2021-11-16, 15:05
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Replies: 7
Views: 624
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...
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Forum: Miscellaneous Math
2021-11-16, 10:11
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Replies: 557
Views: 64,887
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Forum: Miscellaneous Math
2021-11-16, 08:49
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Replies: 7
Views: 624
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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Forum: Miscellaneous Math
2021-11-16, 08:23
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Replies: 557
Views: 64,887
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Forum: Miscellaneous Math
2021-11-11, 20:32
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Replies: 557
Views: 64,887
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Forum: Miscellaneous Math
2021-10-13, 15:15
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Replies: 52
Views: 4,054
Indeed. His omniscience suggests that it is...
Indeed. His omniscience suggests that it is necessary to substantially limit both omnipotence and omniscience. and then holy Bible marasmus and mersenneforum are not far away
I asked my neighbor...
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Forum: Miscellaneous Math
2021-10-12, 17:29
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Replies: 52
Views: 4,054
***
God is almighty. ok. Let God create a BIG Stone, that is too big to god can't lift by blessed god arms, ok Seems that God is not too Almighty??? No. it's obviously. He just do not do such things.
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Forum: Miscellaneous Math
2021-08-26, 15:01
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Replies: 0
Views: 439
Inverse of FFT multiplication?
on the sample
101*109=11009 (1)
let x=10
x^4+x^3+9
at the same time
(x^2+1)*(x^2+9)=x^4+10*x^2+9
i.e.
x^4+x^3+9-(x^3-10*x^2)=x^4+10*x^2+9
Bold polinomial exist for any composite.
So if we...
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Forum: Dobri
2021-08-26, 13:25
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Replies: 53
Views: 4,648
100+ years old talk to Doctor
-friend of...
100+ years old talk to Doctor
-friend of mine tell me, that he can have sex in row at the one night, and I can not, Help me!
-Ok, its very easy to Help! You just can tell him that You can do the...
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Forum: Miscellaneous Math
2021-08-20, 14:57
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Replies: 45
Views: 4,692
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Forum: Miscellaneous Math
2021-07-31, 15:19
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Replies: 20
Views: 2,127
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Forum: Miscellaneous Math
2021-07-30, 18:22
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Replies: 20
Views: 2,127
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Forum: Miscellaneous Math
2021-07-30, 10:30
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Replies: 20
Views: 2,127
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Forum: Miscellaneous Math
2021-07-29, 19:24
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Replies: 20
Views: 2,127
) we start from (b-y)^2? For cube, (b-y)^3 must...
) we start from (b-y)^2? For cube, (b-y)^3 must be expanded (b^3-3*b^2*y+3*b*y^2-y^3) and solved in whole by the same manner, without any simplification from the start. And I'm stuck with imaginary...
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