Is this generalization of CRT true?

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$a^n equiv b_1 bmod p_1$



$a^n equiv b_2 bmod p_2$



if $gcd(b_1,p_2) = gcd(b_2,p_1) = 1$ then $a^n equiv b_1 cdot b_2 bmod p_1cdot p_2$



$p_1$ and $p_2$ are not necessarily coprime.



edit



what about $gcd(p_1, p_2) = 1$ ?




discrete-mathematics




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  • 1




    $a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
    – user578878
    Jul 29 at 16:12






  • 1




    To create a subscript, use the underscore. Now p_2 becomes $p_2$.
    – hardmath
    Jul 29 at 16:13






  • 1




    Same example as above.
    – user578878
    Jul 29 at 16:18










  • @nextpuzzle thanks
    – hjx
    Jul 29 at 16:20










  • Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
    – Joffan
    Jul 29 at 16:37














up vote
0
down vote

favorite












$a^n equiv b_1 bmod p_1$



$a^n equiv b_2 bmod p_2$



if $gcd(b_1,p_2) = gcd(b_2,p_1) = 1$ then $a^n equiv b_1 cdot b_2 bmod p_1cdot p_2$



$p_1$ and $p_2$ are not necessarily coprime.



edit



what about $gcd(p_1, p_2) = 1$ ?




discrete-mathematics




share|cite|improve this question

















  • 1




    $a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
    – user578878
    Jul 29 at 16:12






  • 1




    To create a subscript, use the underscore. Now p_2 becomes $p_2$.
    – hardmath
    Jul 29 at 16:13






  • 1




    Same example as above.
    – user578878
    Jul 29 at 16:18










  • @nextpuzzle thanks
    – hjx
    Jul 29 at 16:20










  • Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
    – Joffan
    Jul 29 at 16:37












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$a^n equiv b_1 bmod p_1$



$a^n equiv b_2 bmod p_2$



if $gcd(b_1,p_2) = gcd(b_2,p_1) = 1$ then $a^n equiv b_1 cdot b_2 bmod p_1cdot p_2$



$p_1$ and $p_2$ are not necessarily coprime.



edit



what about $gcd(p_1, p_2) = 1$ ?




discrete-mathematics




share|cite|improve this question













$a^n equiv b_1 bmod p_1$



$a^n equiv b_2 bmod p_2$



if $gcd(b_1,p_2) = gcd(b_2,p_1) = 1$ then $a^n equiv b_1 cdot b_2 bmod p_1cdot p_2$



$p_1$ and $p_2$ are not necessarily coprime.



edit



what about $gcd(p_1, p_2) = 1$ ?





discrete-mathematics





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 16:14
























asked Jul 29 at 16:06









hjx

36719




36719







  • 1




    $a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
    – user578878
    Jul 29 at 16:12






  • 1




    To create a subscript, use the underscore. Now p_2 becomes $p_2$.
    – hardmath
    Jul 29 at 16:13






  • 1




    Same example as above.
    – user578878
    Jul 29 at 16:18










  • @nextpuzzle thanks
    – hjx
    Jul 29 at 16:20










  • Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
    – Joffan
    Jul 29 at 16:37












  • 1




    $a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
    – user578878
    Jul 29 at 16:12






  • 1




    To create a subscript, use the underscore. Now p_2 becomes $p_2$.
    – hardmath
    Jul 29 at 16:13






  • 1




    Same example as above.
    – user578878
    Jul 29 at 16:18










  • @nextpuzzle thanks
    – hjx
    Jul 29 at 16:20










  • Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
    – Joffan
    Jul 29 at 16:37







1




1




$a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
– user578878
Jul 29 at 16:12




$a=5$, $n=1$, $b_1=2$, $p_1=3$, $b_2=3$, $p_2=2$, but $a^n=5notequiv 6=b_1b_2pmodp_1p_2=6$.
– user578878
Jul 29 at 16:12




1




1




To create a subscript, use the underscore. Now p_2 becomes $p_2$.
– hardmath
Jul 29 at 16:13




To create a subscript, use the underscore. Now p_2 becomes $p_2$.
– hardmath
Jul 29 at 16:13




1




1




Same example as above.
– user578878
Jul 29 at 16:18




Same example as above.
– user578878
Jul 29 at 16:18












@nextpuzzle thanks
– hjx
Jul 29 at 16:20




@nextpuzzle thanks
– hjx
Jul 29 at 16:20












Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
– Joffan
Jul 29 at 16:37




Note that you do not solve CRT simultaneous equivalences by multiplication, in general.
– Joffan
Jul 29 at 16:37















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