Mixing
Prove that $ a^{[phi(m), phi(n)]} equiv 1 pmod{mn} $
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Given that $m,n > 2$ are relatively prime integers and that $a$ is an integer relatively prime to $mn$, prove that
$$ a^{[phi(m), phi(n)]}equiv 1 pmod{mn} $$
I started by using the fact that
$$ [phi(m), phi(n)] = alphaphi(m)=betaphi(n) $$
for some positive integers $ alpha , beta $ to then write
$$ a^{alphaphi(m)}=(a^{phi(m)})^alphaequiv(1)^alphaequiv1 pmod{m} $$
and
$$ a^{betaphi(n)}=(a^{phi(m)})^alphaequiv(1)^betaequiv1 pmod{n} $$
I'm wondering if what I did was correct and how to apply the Chinese Remainder Theorem to show that this congruence is true mod mn.
elementary-number-theory proof-verification totient-function
edited yesterday
Lynn
2,5901526
asked yesterday
mjoseph
477
add a comment |
up vote
1
down vote
favorite
Given that $m,n > 2$ are relatively prime integers and that $a$ is an integer relatively prime to $mn$, prove that
$$ a^{[phi(m), phi(n)]}equiv 1 pmod{mn} $$
I started by using the fact that
$$ [phi(m), phi(n)] = alphaphi(m)=betaphi(n) $$
for some positive integers $ alpha , beta $ to then write
$$ a^{alphaphi(m)}=(a^{phi(m)})^alphaequiv(1)^alphaequiv1 pmod{m} $$
and
$$ a^{betaphi(n)}=(a^{phi(m)})^alphaequiv(1)^betaequiv1 pmod{n} $$
I'm wondering if what I did was correct and how to apply the Chinese Remainder Theorem to show that this congruence is true mod mn.
elementary-number-theory proof-verification totient-function
edited yesterday
Lynn
2,5901526
asked yesterday
mjoseph
477
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given that $m,n > 2$ are relatively prime integers and that $a$ is an integer relatively prime to $mn$, prove that
$$ a^{[phi(m), phi(n)]}equiv 1 pmod{mn} $$
I started by using the fact that
$$ [phi(m), phi(n)] = alphaphi(m)=betaphi(n) $$
for some positive integers $ alpha , beta $ to then write
$$ a^{alphaphi(m)}=(a^{phi(m)})^alphaequiv(1)^alphaequiv1 pmod{m} $$
and
$$ a^{betaphi(n)}=(a^{phi(m)})^alphaequiv(1)^betaequiv1 pmod{n} $$
I'm wondering if what I did was correct and how to apply the Chinese Remainder Theorem to show that this congruence is true mod mn.
elementary-number-theory proof-verification totient-function
edited yesterday
Lynn
2,5901526
asked yesterday
mjoseph
477
Given that $m,n > 2$ are relatively prime integers and that $a$ is an integer relatively prime to $mn$, prove that
$$ a^{[phi(m), phi(n)]}equiv 1 pmod{mn} $$
I started by using the fact that
$$ [phi(m), phi(n)] = alphaphi(m)=betaphi(n) $$
for some positive integers $ alpha , beta $ to then write
$$ a^{alphaphi(m)}=(a^{phi(m)})^alphaequiv(1)^alphaequiv1 pmod{m} $$
and
$$ a^{betaphi(n)}=(a^{phi(m)})^alphaequiv(1)^betaequiv1 pmod{n} $$
I'm wondering if what I did was correct and how to apply the Chinese Remainder Theorem to show that this congruence is true mod mn.
elementary-number-theory proof-verification totient-function
elementary-number-theory proof-verification totient-function
edited yesterday
Lynn
2,5901526
asked yesterday
mjoseph
477
edited yesterday
Lynn
2,5901526
asked yesterday
mjoseph
477
edited yesterday
Lynn
2,5901526
edited yesterday
Lynn
2,5901526
edited yesterday
Lynn
2,5901526
2,5901526
asked yesterday
mjoseph
477
asked yesterday
mjoseph
477
asked yesterday
mjoseph
477
477
add a comment |
add a comment |
1 Answer
1
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up vote
1
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Yes, it is correct. To finish, by CRT: $,Aequiv 1bmod m & niff Aequiv 1pmod{!mn}.,$ Or, w/o CRT, we have $,m,nmid A-1iff {rm lcm}(m,n)mid A-1,,$ and $,{rm lcm}(m,n) = mn,$ by $,gcd(m,n)=1$
Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem
answered yesterday
Bill Dubuque
206k29189621
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, it is correct. To finish, by CRT: $,Aequiv 1bmod m & niff Aequiv 1pmod{!mn}.,$ Or, w/o CRT, we have $,m,nmid A-1iff {rm lcm}(m,n)mid A-1,,$ and $,{rm lcm}(m,n) = mn,$ by $,gcd(m,n)=1$
Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem
answered yesterday
Bill Dubuque
206k29189621
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
add a comment |
up vote
1
down vote
accepted
Yes, it is correct. To finish, by CRT: $,Aequiv 1bmod m & niff Aequiv 1pmod{!mn}.,$ Or, w/o CRT, we have $,m,nmid A-1iff {rm lcm}(m,n)mid A-1,,$ and $,{rm lcm}(m,n) = mn,$ by $,gcd(m,n)=1$
Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem
answered yesterday
Bill Dubuque
206k29189621
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, it is correct. To finish, by CRT: $,Aequiv 1bmod m & niff Aequiv 1pmod{!mn}.,$ Or, w/o CRT, we have $,m,nmid A-1iff {rm lcm}(m,n)mid A-1,,$ and $,{rm lcm}(m,n) = mn,$ by $,gcd(m,n)=1$
Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem
answered yesterday
Bill Dubuque
206k29189621
Yes, it is correct. To finish, by CRT: $,Aequiv 1bmod m & niff Aequiv 1pmod{!mn}.,$ Or, w/o CRT, we have $,m,nmid A-1iff {rm lcm}(m,n)mid A-1,,$ and $,{rm lcm}(m,n) = mn,$ by $,gcd(m,n)=1$
Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem
answered yesterday
Bill Dubuque
206k29189621
answered yesterday
Bill Dubuque
206k29189621
answered yesterday
Bill Dubuque
206k29189621
answered yesterday
Bill Dubuque
206k29189621
206k29189621
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
add a comment |
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
what is A? and how did you get it using the CRT?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case?
– mjoseph
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
yesterday
add a comment |
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