Prove there exist $a$ such that $ab pmod p = 1$, where $p$ is a prime number and $bin {0,1,2,…p−1}$
$begingroup$
How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number
$$ab pmod p = 1$$
number-theory elementary-number-theory
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
$endgroup$
add a comment |
$begingroup$
How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number
$$ab pmod p = 1$$
number-theory elementary-number-theory
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
$endgroup$
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22
add a comment |
$begingroup$
How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number
$$ab pmod p = 1$$
number-theory elementary-number-theory
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
$endgroup$
How do I prove there exist a modulo inverse of $b$, lets call it $a$, where $b in {0,1,2,....p−1} $ and $p$ is a prime number
$$ab pmod p = 1$$
number-theory elementary-number-theory
number-theory elementary-number-theory
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
edited Dec 19 '18 at 20:40
rtybase
11.5k31534
11.5k31534
asked Oct 21 '18 at 2:06
CupCup
326
asked Oct 21 '18 at 2:06
CupCup
326
asked Oct 21 '18 at 2:06
CupCup
326
326
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22
add a comment |
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as
Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$
E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.
The proof of this theorem is left as an exercise.
Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.
answered Oct 21 '18 at 2:18
TreborTrebor
99815
$endgroup$
add a comment |
$begingroup$
You need to exclude the case $b=0$.
Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
$$b^{p-1}equiv 1 pmod{p}$$
or
$$bcdot b^{p-2}equiv 1 pmod{p}$$
and $a = b^{p-2} pmod{p}$.
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as
Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$
E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.
The proof of this theorem is left as an exercise.
Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.
answered Oct 21 '18 at 2:18
TreborTrebor
99815
$endgroup$
add a comment |
$begingroup$
Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as
Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$
E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.
The proof of this theorem is left as an exercise.
Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.
answered Oct 21 '18 at 2:18
TreborTrebor
99815
$endgroup$
add a comment |
$begingroup$
Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as
Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$
E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.
The proof of this theorem is left as an exercise.
Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.
answered Oct 21 '18 at 2:18
TreborTrebor
99815
$endgroup$
Hint: The statement you want to prove is equivalent to finding
$$
ab+kp=1,
$$
for some $k inmathbb Z$. This is called Bézout's identity, which can be stated more precisely as
Theorem (Bézout). For all positive integers $u,v$ with $gcd(u,v)=1$, there exists integers $m,n$ such that $$mu+nv=1.$$
E.g. $gcd(17,6)=1$ implies that there are some integers $m,n$ such that $17m+6n=1$. One of the solution is $m=1,n=-3$.
The proof of this theorem is left as an exercise.
Now as $gcd(a, p)=1$, there exists $b,k$ such that $ab+kp=1$ holds. However, you need to show that $b$ can always be found in the specified range, ${0,1,dots,p-1}$. This is also quite easy.
answered Oct 21 '18 at 2:18
TreborTrebor
99815
answered Oct 21 '18 at 2:18
TreborTrebor
99815
answered Oct 21 '18 at 2:18
TreborTrebor
99815
answered Oct 21 '18 at 2:18
TreborTrebor
99815
99815
add a comment |
add a comment |
$begingroup$
You need to exclude the case $b=0$.
Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
You need to exclude the case $b=0$.
Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
You need to exclude the case $b=0$.
Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
$endgroup$
You need to exclude the case $b=0$.
Use the Pigeonhole Principle together with the fact that multiplication by $b mod p$ is one-to-one.
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
answered Oct 21 '18 at 3:53
Robert IsraelRobert Israel
330k23219473
330k23219473
add a comment |
add a comment |
$begingroup$
Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
$$b^{p-1}equiv 1 pmod{p}$$
or
$$bcdot b^{p-2}equiv 1 pmod{p}$$
and $a = b^{p-2} pmod{p}$.
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
$endgroup$
add a comment |
$begingroup$
Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
$$b^{p-1}equiv 1 pmod{p}$$
or
$$bcdot b^{p-2}equiv 1 pmod{p}$$
and $a = b^{p-2} pmod{p}$.
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
$endgroup$
add a comment |
$begingroup$
Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
$$b^{p-1}equiv 1 pmod{p}$$
or
$$bcdot b^{p-2}equiv 1 pmod{p}$$
and $a = b^{p-2} pmod{p}$.
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
$endgroup$
Alternatively, you can use FLT. Because $gcd(b,p)=1, forall bin {1,...,p-1}, p>2$, we have
$$b^{p-1}equiv 1 pmod{p}$$
or
$$bcdot b^{p-2}equiv 1 pmod{p}$$
and $a = b^{p-2} pmod{p}$.
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
answered Oct 21 '18 at 10:00
rtybasertybase
11.5k31534
11.5k31534
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/B%C3%A9zout%27s_identity
$endgroup$
– Lord Shark the Unknown
Oct 21 '18 at 2:08
$begingroup$
@LordSharktheUnknown im sorry but I am really stupid. How does this identity help me solve this question?
$endgroup$
– Cup
Oct 21 '18 at 2:17
$begingroup$
There is no inverse if $b = 0$.
$endgroup$
– JavaMan
Oct 21 '18 at 2:22