Mixing
How many solutions does $x equiv x^{-1} pmod n$ have?
$begingroup$
How many solutions does $x equiv x^{-1} pmod n$ have?
$n$ is defined to be a positive integer,
What I believe the solution will be is along the lines of 2 cases:
When $n = 1$, the set of solutions will just be $x in mathbb Z$ because $x equiv x^{-1} pmod 1$ can be rewritten as $x - x^{-1} equiv 0 pmod 1$ and mod 1 of any integer will always reduce to 0
When $n gt 1$, I know we have to apply Chinese Remainder Theorem somehow, I just do not know how to approach this though. We can claim by the Fundamental Theorem of Arithmetic that $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{l}^{k_{l}}$ where $p_{i}$ are prime numbers. How would I continue?$\$
I thought maybe multiplying both sides by $x$ and rearranging to get $x^{2} equiv 1pmod n$ could be of some help but I couldn't get any further.
elementary-number-theory prime-numbers chinese-remainder-theorem
edited Jan 18 at 23:43
Bernard
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
$endgroup$
|
show 2 more comments
$begingroup$
How many solutions does $x equiv x^{-1} pmod n$ have?
$n$ is defined to be a positive integer,
What I believe the solution will be is along the lines of 2 cases:
When $n = 1$, the set of solutions will just be $x in mathbb Z$ because $x equiv x^{-1} pmod 1$ can be rewritten as $x - x^{-1} equiv 0 pmod 1$ and mod 1 of any integer will always reduce to 0
When $n gt 1$, I know we have to apply Chinese Remainder Theorem somehow, I just do not know how to approach this though. We can claim by the Fundamental Theorem of Arithmetic that $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{l}^{k_{l}}$ where $p_{i}$ are prime numbers. How would I continue?$\$
I thought maybe multiplying both sides by $x$ and rearranging to get $x^{2} equiv 1pmod n$ could be of some help but I couldn't get any further.
elementary-number-theory prime-numbers chinese-remainder-theorem
edited Jan 18 at 23:43
Bernard
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
$endgroup$
2
$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
1
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01
|
show 2 more comments
$begingroup$
How many solutions does $x equiv x^{-1} pmod n$ have?
$n$ is defined to be a positive integer,
What I believe the solution will be is along the lines of 2 cases:
When $n = 1$, the set of solutions will just be $x in mathbb Z$ because $x equiv x^{-1} pmod 1$ can be rewritten as $x - x^{-1} equiv 0 pmod 1$ and mod 1 of any integer will always reduce to 0
When $n gt 1$, I know we have to apply Chinese Remainder Theorem somehow, I just do not know how to approach this though. We can claim by the Fundamental Theorem of Arithmetic that $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{l}^{k_{l}}$ where $p_{i}$ are prime numbers. How would I continue?$\$
I thought maybe multiplying both sides by $x$ and rearranging to get $x^{2} equiv 1pmod n$ could be of some help but I couldn't get any further.
elementary-number-theory prime-numbers chinese-remainder-theorem
edited Jan 18 at 23:43
Bernard
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
$endgroup$
How many solutions does $x equiv x^{-1} pmod n$ have?
$n$ is defined to be a positive integer,
What I believe the solution will be is along the lines of 2 cases:
When $n = 1$, the set of solutions will just be $x in mathbb Z$ because $x equiv x^{-1} pmod 1$ can be rewritten as $x - x^{-1} equiv 0 pmod 1$ and mod 1 of any integer will always reduce to 0
When $n gt 1$, I know we have to apply Chinese Remainder Theorem somehow, I just do not know how to approach this though. We can claim by the Fundamental Theorem of Arithmetic that $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{l}^{k_{l}}$ where $p_{i}$ are prime numbers. How would I continue?$\$
I thought maybe multiplying both sides by $x$ and rearranging to get $x^{2} equiv 1pmod n$ could be of some help but I couldn't get any further.
elementary-number-theory prime-numbers chinese-remainder-theorem
elementary-number-theory prime-numbers chinese-remainder-theorem
edited Jan 18 at 23:43
Bernard
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
edited Jan 18 at 23:43
Bernard
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
edited Jan 18 at 23:43
Bernard
121k740116
edited Jan 18 at 23:43
Bernard
121k740116
edited Jan 18 at 23:43
Bernard
121k740116
121k740116
asked Jan 18 at 23:29
WallaceWallace
1367
asked Jan 18 at 23:29
WallaceWallace
1367
asked Jan 18 at 23:29
WallaceWallace
1367
1367
2
$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
1
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01
|
show 2 more comments
2
$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
1
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01
2
2
$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
1
1
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01
|
show 2 more comments
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$begingroup$
$x equiv x^{-1}$ is equivalent to $x^2 - 1 equiv (x+1)(x-1) equiv 0$.
$endgroup$
– 0x539
Jan 18 at 23:57
$begingroup$
Hint: $(x!-!1,x!+!1) = (x!-!1,2)$ is coprime to odd primes $p,,$ so $,p^nmid (x!-!1)(x!+!1),Rightarrow,p^nmid x-1$ or $,p^nmid x+1 $
$endgroup$
– Bill Dubuque
Jan 19 at 0:28
$begingroup$
@0x539 how does that help? I get that part but I do not understand where to go from that. I actually went down that path but I didn't know how to pursue from there, would we could consider separate cases for $(x+1)$ and $(x-1)$?
$endgroup$
– Wallace
Jan 19 at 0:32
$begingroup$
@Wallace First if $n$ is prime it shows that the only solution are $1, -1$. If $n$ is composite then $x+1$ and $x-1$ need to contain all prime factors of $n$. This helps somewhat, for example if $n$ is an odd prime power then the only solutions are again $1, -1$. I don't claim this will provide a full solution but it seems promising to me.
$endgroup$
– 0x539
Jan 19 at 0:40
1
$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
Jan 20 at 17:01