Mixing
Showing two different moduli are same
$begingroup$
$t$ is an integer coprime to primes $p,a,b$.
$max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.
$a,bin[p^{1/4},2p^{1/4}]$ holds.
$t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$
Denote $r=tabbmod pb$.
There are two ways to interpret $r$.
$tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.
$tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$
How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?
elementary-number-theory modular-arithmetic
asked Jan 15 at 19:27
BroutBrout
2,5591431
$endgroup$
add a comment |
$begingroup$
$t$ is an integer coprime to primes $p,a,b$.
$max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.
$a,bin[p^{1/4},2p^{1/4}]$ holds.
$t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$
Denote $r=tabbmod pb$.
There are two ways to interpret $r$.
$tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.
$tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$
How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?
elementary-number-theory modular-arithmetic
asked Jan 15 at 19:27
BroutBrout
2,5591431
$endgroup$
add a comment |
$begingroup$
$t$ is an integer coprime to primes $p,a,b$.
$max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.
$a,bin[p^{1/4},2p^{1/4}]$ holds.
$t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$
Denote $r=tabbmod pb$.
There are two ways to interpret $r$.
$tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.
$tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$
How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?
elementary-number-theory modular-arithmetic
asked Jan 15 at 19:27
BroutBrout
2,5591431
$endgroup$
$t$ is an integer coprime to primes $p,a,b$.
$max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.
$a,bin[p^{1/4},2p^{1/4}]$ holds.
$t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$
Denote $r=tabbmod pb$.
There are two ways to interpret $r$.
$tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.
$tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$
How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked Jan 15 at 19:27
BroutBrout
2,5591431
asked Jan 15 at 19:27
BroutBrout
2,5591431
edited Jan 15 at 19:59
Brout
asked Jan 15 at 19:27
BroutBrout
2,5591431
asked Jan 15 at 19:27
BroutBrout
2,5591431
asked Jan 15 at 19:27
BroutBrout
2,5591431
2,5591431
add a comment |
add a comment |
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