Mixing
Modular Exponentiation with unknown base
$begingroup$
Consider $x^aequiv1 pmod n$.
Is there a general way to solve for $x$, given $a$ and $n$?
Would knowing the factorization of $n$ make it easier?
modular-arithmetic
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
$endgroup$
add a comment |
$begingroup$
Consider $x^aequiv1 pmod n$.
Is there a general way to solve for $x$, given $a$ and $n$?
Would knowing the factorization of $n$ make it easier?
modular-arithmetic
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
$endgroup$
add a comment |
$begingroup$
Consider $x^aequiv1 pmod n$.
Is there a general way to solve for $x$, given $a$ and $n$?
Would knowing the factorization of $n$ make it easier?
modular-arithmetic
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
$endgroup$
Consider $x^aequiv1 pmod n$.
Is there a general way to solve for $x$, given $a$ and $n$?
Would knowing the factorization of $n$ make it easier?
modular-arithmetic
modular-arithmetic
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
asked Jan 23 at 9:03
Ruan SunkelRuan Sunkel
385
385
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $n=p_1^{e_1}cdots p_r^{e_r}$ is the factorization of $n$ into prime powers, then you can solve simultaneously (by the Chinese remainder theorem)
$$x^aequiv 1mod p_i^{e_i},quad 1leq ileq r,$$
and then assemble the solutions together. This solution is unique in the range of $0$ to $n-1$.
In particular, if $e_i=1$, then you can use Fermat's little theorem, $x^{p_i-1}equiv 1mod p_i$ to simplify the associated congruence.
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
$endgroup$
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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votes
$begingroup$
If $n=p_1^{e_1}cdots p_r^{e_r}$ is the factorization of $n$ into prime powers, then you can solve simultaneously (by the Chinese remainder theorem)
$$x^aequiv 1mod p_i^{e_i},quad 1leq ileq r,$$
and then assemble the solutions together. This solution is unique in the range of $0$ to $n-1$.
In particular, if $e_i=1$, then you can use Fermat's little theorem, $x^{p_i-1}equiv 1mod p_i$ to simplify the associated congruence.
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
$endgroup$
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
add a comment |
$begingroup$
If $n=p_1^{e_1}cdots p_r^{e_r}$ is the factorization of $n$ into prime powers, then you can solve simultaneously (by the Chinese remainder theorem)
$$x^aequiv 1mod p_i^{e_i},quad 1leq ileq r,$$
and then assemble the solutions together. This solution is unique in the range of $0$ to $n-1$.
In particular, if $e_i=1$, then you can use Fermat's little theorem, $x^{p_i-1}equiv 1mod p_i$ to simplify the associated congruence.
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
$endgroup$
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
add a comment |
$begingroup$
If $n=p_1^{e_1}cdots p_r^{e_r}$ is the factorization of $n$ into prime powers, then you can solve simultaneously (by the Chinese remainder theorem)
$$x^aequiv 1mod p_i^{e_i},quad 1leq ileq r,$$
and then assemble the solutions together. This solution is unique in the range of $0$ to $n-1$.
In particular, if $e_i=1$, then you can use Fermat's little theorem, $x^{p_i-1}equiv 1mod p_i$ to simplify the associated congruence.
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
$endgroup$
If $n=p_1^{e_1}cdots p_r^{e_r}$ is the factorization of $n$ into prime powers, then you can solve simultaneously (by the Chinese remainder theorem)
$$x^aequiv 1mod p_i^{e_i},quad 1leq ileq r,$$
and then assemble the solutions together. This solution is unique in the range of $0$ to $n-1$.
In particular, if $e_i=1$, then you can use Fermat's little theorem, $x^{p_i-1}equiv 1mod p_i$ to simplify the associated congruence.
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
answered Jan 23 at 9:15
WuestenfuxWuestenfux
4,9921513
4,9921513
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
add a comment |
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
For $e_i > 1$, we can use $x^{phi(p_i^{e_i})} equiv 1 mod{p_i^{e_i}}$, where $phi(p_i^{e_i}) = p_i^{e_i}-p_i^{e_i-1}$.
$endgroup$
– Dirk Liebhold
Jan 23 at 9:37
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Thanks guys, may I ask for a simple example?
$endgroup$
– Ruan Sunkel
Jan 23 at 9:40
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
$begingroup$
Dirk: Indeed, the specific congruence $x^aequiv 1 mod n$ allows more to say.
$endgroup$
– Wuestenfux
Jan 23 at 9:44
add a comment |
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