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If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$. [duplicate]
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For any prime $p > 3$, why is $p^2-1$ always divisible by 24?
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If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$.
linear-algebra abstract-algebra
edited 2 days ago
David G. Stork
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marked as duplicate by Théophile, T. Bongers, Bill Dubuque abstract-algebra
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
For any prime $p > 3$, why is $p^2-1$ always divisible by 24?
16 answers
If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$.
linear-algebra abstract-algebra
edited 2 days ago
David G. Stork
8,97421232
asked 2 days ago
ten1o
1335
marked as duplicate by Théophile, T. Bongers, Bill Dubuque abstract-algebra
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This question already has an answer here:
For any prime $p > 3$, why is $p^2-1$ always divisible by 24?
16 answers
If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$.
linear-algebra abstract-algebra
edited 2 days ago
David G. Stork
8,97421232
asked 2 days ago
ten1o
1335
This question already has an answer here:
For any prime $p > 3$, why is $p^2-1$ always divisible by 24?
16 answers
If $(6,n)=1$, prove that $n^2-1$ is divisible by $24$.
This question already has an answer here:
For any prime $p > 3$, why is $p^2-1$ always divisible by 24?
16 answers
linear-algebra abstract-algebra
linear-algebra abstract-algebra
edited 2 days ago
David G. Stork
8,97421232
asked 2 days ago
ten1o
1335
edited 2 days ago
David G. Stork
8,97421232
asked 2 days ago
ten1o
1335
edited 2 days ago
David G. Stork
8,97421232
edited 2 days ago
David G. Stork
8,97421232
edited 2 days ago
David G. Stork
8,97421232
8,97421232
asked 2 days ago
ten1o
1335
asked 2 days ago
ten1o
1335
asked 2 days ago
ten1o
1335
1335
marked as duplicate by Théophile, T. Bongers, Bill Dubuque abstract-algebra
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Théophile, T. Bongers, Bill Dubuque abstract-algebra
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago
add a comment |
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago
add a comment |
3 Answers
3
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up vote
2
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Hint
$$(6,n)=1implies$$
$ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
answered 2 days ago
hamam_Abdallah
36.5k21533
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
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up vote
0
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If $(6,n) = 1,$ then $n equiv 1 pmod 6$ or $n equiv 5 pmod 6,$ since $varphi(6) = 2.$
So, we have two cases:
► $n equiv 1 pmod 6:$
$$n equiv 1 pmod 6 Rightarrow n^2 equiv 1^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
► $n equiv 5 pmod 6 Rightarrow n equiv -1 pmod 6:$
$$n equiv -1 pmod 6 Rightarrow n^2 equiv (-1)^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
In both, we conclude that $n^2 - 1 equiv 0 pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
answered 2 days ago
674123173797 - 4
1237
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$(6,n)=1$ $implies$ $n$ is odd.
Therefore , $$n^{2}equiv 1 pmod8$$
Also $(3,n)=1$ , since $(2times3,n)=1$
Therefore by Fermat's theorem ,$$n^2equiv 1pmod3$$.
Now you can directly combine these two congruence, since $(3,8)=1$
Therefore, $$n^2equiv 1pmod{24}$$
answered 2 days ago
Samurai
987
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint
$$(6,n)=1implies$$
$ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
answered 2 days ago
hamam_Abdallah
36.5k21533
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
add a comment |
up vote
2
down vote
Hint
$$(6,n)=1implies$$
$ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
answered 2 days ago
hamam_Abdallah
36.5k21533
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint
$$(6,n)=1implies$$
$ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
answered 2 days ago
hamam_Abdallah
36.5k21533
Hint
$$(6,n)=1implies$$
$ n=6k+1 $ or $n=6k-1$.
in the first case
$$n^2-1=36k^2+12k=12k(3k+1)$$
in the second
$$n^2-1=12k(3k-1).$$
If $k$ is even ....
answered 2 days ago
hamam_Abdallah
36.5k21533
answered 2 days ago
hamam_Abdallah
36.5k21533
answered 2 days ago
hamam_Abdallah
36.5k21533
answered 2 days ago
hamam_Abdallah
36.5k21533
36.5k21533
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
add a comment |
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
math.stackexchange.com/questions/507451/suppose-that-p-≥-q-≥-5-are-both-prime-numbers-prove-that-24-divides-p2/507511#507511
– lab bhattacharjee
2 days ago
add a comment |
up vote
0
down vote
If $(6,n) = 1,$ then $n equiv 1 pmod 6$ or $n equiv 5 pmod 6,$ since $varphi(6) = 2.$
So, we have two cases:
► $n equiv 1 pmod 6:$
$$n equiv 1 pmod 6 Rightarrow n^2 equiv 1^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
► $n equiv 5 pmod 6 Rightarrow n equiv -1 pmod 6:$
$$n equiv -1 pmod 6 Rightarrow n^2 equiv (-1)^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
In both, we conclude that $n^2 - 1 equiv 0 pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
answered 2 days ago
674123173797 - 4
1237
add a comment |
up vote
0
down vote
If $(6,n) = 1,$ then $n equiv 1 pmod 6$ or $n equiv 5 pmod 6,$ since $varphi(6) = 2.$
So, we have two cases:
► $n equiv 1 pmod 6:$
$$n equiv 1 pmod 6 Rightarrow n^2 equiv 1^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
► $n equiv 5 pmod 6 Rightarrow n equiv -1 pmod 6:$
$$n equiv -1 pmod 6 Rightarrow n^2 equiv (-1)^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
In both, we conclude that $n^2 - 1 equiv 0 pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
answered 2 days ago
674123173797 - 4
1237
add a comment |
up vote
0
down vote
up vote
0
down vote
If $(6,n) = 1,$ then $n equiv 1 pmod 6$ or $n equiv 5 pmod 6,$ since $varphi(6) = 2.$
So, we have two cases:
► $n equiv 1 pmod 6:$
$$n equiv 1 pmod 6 Rightarrow n^2 equiv 1^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
► $n equiv 5 pmod 6 Rightarrow n equiv -1 pmod 6:$
$$n equiv -1 pmod 6 Rightarrow n^2 equiv (-1)^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
In both, we conclude that $n^2 - 1 equiv 0 pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
answered 2 days ago
674123173797 - 4
1237
If $(6,n) = 1,$ then $n equiv 1 pmod 6$ or $n equiv 5 pmod 6,$ since $varphi(6) = 2.$
So, we have two cases:
► $n equiv 1 pmod 6:$
$$n equiv 1 pmod 6 Rightarrow n^2 equiv 1^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
► $n equiv 5 pmod 6 Rightarrow n equiv -1 pmod 6:$
$$n equiv -1 pmod 6 Rightarrow n^2 equiv (-1)^2 pmod 6 Rightarrow n^2 - 1 equiv 0 pmod 6 Rightarrow boxed{n^2 - 1 equiv 0 pmod{24}}$$
In both, we conclude that $n^2 - 1 equiv 0 pmod{24}.$ Therefore, $n^2 - 1$ is divisible by $24$ when $(6,n) = 1.$
answered 2 days ago
674123173797 - 4
1237
answered 2 days ago
674123173797 - 4
1237
answered 2 days ago
674123173797 - 4
1237
answered 2 days ago
674123173797 - 4
1237
1237
add a comment |
add a comment |
up vote
0
down vote
$(6,n)=1$ $implies$ $n$ is odd.
Therefore , $$n^{2}equiv 1 pmod8$$
Also $(3,n)=1$ , since $(2times3,n)=1$
Therefore by Fermat's theorem ,$$n^2equiv 1pmod3$$.
Now you can directly combine these two congruence, since $(3,8)=1$
Therefore, $$n^2equiv 1pmod{24}$$
answered 2 days ago
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
$(6,n)=1$ $implies$ $n$ is odd.
Therefore , $$n^{2}equiv 1 pmod8$$
Also $(3,n)=1$ , since $(2times3,n)=1$
Therefore by Fermat's theorem ,$$n^2equiv 1pmod3$$.
Now you can directly combine these two congruence, since $(3,8)=1$
Therefore, $$n^2equiv 1pmod{24}$$
answered 2 days ago
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
$(6,n)=1$ $implies$ $n$ is odd.
Therefore , $$n^{2}equiv 1 pmod8$$
Also $(3,n)=1$ , since $(2times3,n)=1$
Therefore by Fermat's theorem ,$$n^2equiv 1pmod3$$.
Now you can directly combine these two congruence, since $(3,8)=1$
Therefore, $$n^2equiv 1pmod{24}$$
answered 2 days ago
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$(6,n)=1$ $implies$ $n$ is odd.
Therefore , $$n^{2}equiv 1 pmod8$$
Also $(3,n)=1$ , since $(2times3,n)=1$
Therefore by Fermat's theorem ,$$n^2equiv 1pmod3$$.
Now you can directly combine these two congruence, since $(3,8)=1$
Therefore, $$n^2equiv 1pmod{24}$$
answered 2 days ago
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Samurai
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Samurai
987
answered 2 days ago
Samurai
987
987
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samurai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Okay, what are your thoughts on this so far? What exactly is your question?
– Théophile
2 days ago
I assume this means $gcd(6,n)=1$?
– Kevin Long
2 days ago
Most (if not all) of the proofs in the dupe apply here (even though the title restricts to primes $> 3) $
– Bill Dubuque
2 days ago