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 Nov 19 at 18:00
Lynn
2,5801526
asked Nov 19 at 17:50
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 Nov 19 at 18:00
Lynn
2,5801526
asked Nov 19 at 17:50
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 Nov 19 at 18:00
Lynn
2,5801526
asked Nov 19 at 17:50
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 Nov 19 at 18:00
Lynn
2,5801526
asked Nov 19 at 17:50
mjoseph
477
edited Nov 19 at 18:00
Lynn
2,5801526
asked Nov 19 at 17:50
mjoseph
477
edited Nov 19 at 18:00
Lynn
2,5801526
edited Nov 19 at 18:00
Lynn
2,5801526
edited Nov 19 at 18:00
Lynn
2,5801526
2,5801526
asked Nov 19 at 17:50
mjoseph
477
asked Nov 19 at 17:50
mjoseph
477
asked Nov 19 at 17:50
mjoseph
477
477
add a comment |
add a comment |
1 Answer
1
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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 Nov 19 at 18:02
Bill Dubuque
207k29189624
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
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 Nov 19 at 18:02
Bill Dubuque
207k29189624
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
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 Nov 19 at 18:02
Bill Dubuque
207k29189624
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
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 Nov 19 at 18:02
Bill Dubuque
207k29189624
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 Nov 19 at 18:02
Bill Dubuque
207k29189624
answered Nov 19 at 18:02
Bill Dubuque
207k29189624
answered Nov 19 at 18:02
Bill Dubuque
207k29189624
answered Nov 19 at 18:02
Bill Dubuque
207k29189624
207k29189624
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
add a comment |
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
what is A? and how did you get it using the CRT?
– mjoseph
Nov 19 at 18:06
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
Nov 19 at 18:07
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
Nov 19 at 18:07
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph $A := a^{large [phi(m),phi(n)]}$. You proved $,Aequiv 1pmod m$ and $,Aequiv 1pmod n$
– Bill Dubuque
Nov 19 at 18:10
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
@mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that).
– Bill Dubuque
Nov 19 at 18:14
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
Just looked at the link a little more and understand it now. Thank you!
– mjoseph
Nov 19 at 19:26
add a comment |
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