Mixing
Solving an equation mod $n$
Part of a solution, I'm trying to calculate $(n-1)^2$ in $mod n$ when $ninmathbb{N}$. What I tried to do:
$$ (n-1)^2=(n^2-2n+1) (mod n)$$
But I'm not sure what to do next. How should I approach this?
modular-arithmetic
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
add a comment |
Part of a solution, I'm trying to calculate $(n-1)^2$ in $mod n$ when $ninmathbb{N}$. What I tried to do:
$$ (n-1)^2=(n^2-2n+1) (mod n)$$
But I'm not sure what to do next. How should I approach this?
modular-arithmetic
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
add a comment |
Part of a solution, I'm trying to calculate $(n-1)^2$ in $mod n$ when $ninmathbb{N}$. What I tried to do:
$$ (n-1)^2=(n^2-2n+1) (mod n)$$
But I'm not sure what to do next. How should I approach this?
modular-arithmetic
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
Part of a solution, I'm trying to calculate $(n-1)^2$ in $mod n$ when $ninmathbb{N}$. What I tried to do:
$$ (n-1)^2=(n^2-2n+1) (mod n)$$
But I'm not sure what to do next. How should I approach this?
modular-arithmetic
modular-arithmetic
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
edited Nov 20 '18 at 22:25
Jean Marie
28.8k41949
28.8k41949
asked Nov 20 '18 at 21:08
vesii
906
asked Nov 20 '18 at 21:08
vesii
906
asked Nov 20 '18 at 21:08
vesii
906
906
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$n-1equiv -1 pmod n$
So $(n-1)^2 equiv (-1)^2 equiv 1pmod n$.
....
Or $n^2= n*n equiv 0 pmod n$. And $-2nequiv (-2)*nequiv 0 pmod n$. ANd $1 equiv 1pmod n$.
So $n^2 - 2n + 1 equiv 0 + 0 + 1 equiv 1 pmod n$.
...
You just need to keep is mind for $aequiv alpha pmod n$ and $b equiv beta pmod n$
That $a pm bequiv alpha pm beta pmod n$ and $ab equiv alpha beta pmod n$ and $maequiv malpha pmod n$ and $a^k equiv alpha^k pmod n$ and that $n equiv 0 pmod n$.
EVERYTHING falls from that exceedingly nicely.
(In other words nearly all arithmetic distributes over modulus, and that $n$ is equivalent to $0$.)
(So $(n-1)^2$ is over modulus then, under $mod n$ arithmetic, the same thing as $(-1)^2$.)
(A few things to watch out for. Division doesn't distribute unless divisor and modulus are relatively prime. And the powers of exponents are not actually arithmetic so they don't distribute.)
edited Nov 20 '18 at 21:27
zwim
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
add a comment |
Since $${n^2}equiv 0pmod n$$ $$-2nequiv 0pmod n$$
You have $${(n-1)^2}equiv ({n^2}-2n+1)equiv 0+0+1equiv1 pmod n$$
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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active
oldest
votes
$n-1equiv -1 pmod n$
So $(n-1)^2 equiv (-1)^2 equiv 1pmod n$.
....
Or $n^2= n*n equiv 0 pmod n$. And $-2nequiv (-2)*nequiv 0 pmod n$. ANd $1 equiv 1pmod n$.
So $n^2 - 2n + 1 equiv 0 + 0 + 1 equiv 1 pmod n$.
...
You just need to keep is mind for $aequiv alpha pmod n$ and $b equiv beta pmod n$
That $a pm bequiv alpha pm beta pmod n$ and $ab equiv alpha beta pmod n$ and $maequiv malpha pmod n$ and $a^k equiv alpha^k pmod n$ and that $n equiv 0 pmod n$.
EVERYTHING falls from that exceedingly nicely.
(In other words nearly all arithmetic distributes over modulus, and that $n$ is equivalent to $0$.)
(So $(n-1)^2$ is over modulus then, under $mod n$ arithmetic, the same thing as $(-1)^2$.)
(A few things to watch out for. Division doesn't distribute unless divisor and modulus are relatively prime. And the powers of exponents are not actually arithmetic so they don't distribute.)
edited Nov 20 '18 at 21:27
zwim
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
add a comment |
$n-1equiv -1 pmod n$
So $(n-1)^2 equiv (-1)^2 equiv 1pmod n$.
....
Or $n^2= n*n equiv 0 pmod n$. And $-2nequiv (-2)*nequiv 0 pmod n$. ANd $1 equiv 1pmod n$.
So $n^2 - 2n + 1 equiv 0 + 0 + 1 equiv 1 pmod n$.
...
You just need to keep is mind for $aequiv alpha pmod n$ and $b equiv beta pmod n$
That $a pm bequiv alpha pm beta pmod n$ and $ab equiv alpha beta pmod n$ and $maequiv malpha pmod n$ and $a^k equiv alpha^k pmod n$ and that $n equiv 0 pmod n$.
EVERYTHING falls from that exceedingly nicely.
(In other words nearly all arithmetic distributes over modulus, and that $n$ is equivalent to $0$.)
(So $(n-1)^2$ is over modulus then, under $mod n$ arithmetic, the same thing as $(-1)^2$.)
(A few things to watch out for. Division doesn't distribute unless divisor and modulus are relatively prime. And the powers of exponents are not actually arithmetic so they don't distribute.)
edited Nov 20 '18 at 21:27
zwim
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
add a comment |
$n-1equiv -1 pmod n$
So $(n-1)^2 equiv (-1)^2 equiv 1pmod n$.
....
Or $n^2= n*n equiv 0 pmod n$. And $-2nequiv (-2)*nequiv 0 pmod n$. ANd $1 equiv 1pmod n$.
So $n^2 - 2n + 1 equiv 0 + 0 + 1 equiv 1 pmod n$.
...
You just need to keep is mind for $aequiv alpha pmod n$ and $b equiv beta pmod n$
That $a pm bequiv alpha pm beta pmod n$ and $ab equiv alpha beta pmod n$ and $maequiv malpha pmod n$ and $a^k equiv alpha^k pmod n$ and that $n equiv 0 pmod n$.
EVERYTHING falls from that exceedingly nicely.
(In other words nearly all arithmetic distributes over modulus, and that $n$ is equivalent to $0$.)
(So $(n-1)^2$ is over modulus then, under $mod n$ arithmetic, the same thing as $(-1)^2$.)
(A few things to watch out for. Division doesn't distribute unless divisor and modulus are relatively prime. And the powers of exponents are not actually arithmetic so they don't distribute.)
edited Nov 20 '18 at 21:27
zwim
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
$n-1equiv -1 pmod n$
So $(n-1)^2 equiv (-1)^2 equiv 1pmod n$.
....
Or $n^2= n*n equiv 0 pmod n$. And $-2nequiv (-2)*nequiv 0 pmod n$. ANd $1 equiv 1pmod n$.
So $n^2 - 2n + 1 equiv 0 + 0 + 1 equiv 1 pmod n$.
...
You just need to keep is mind for $aequiv alpha pmod n$ and $b equiv beta pmod n$
That $a pm bequiv alpha pm beta pmod n$ and $ab equiv alpha beta pmod n$ and $maequiv malpha pmod n$ and $a^k equiv alpha^k pmod n$ and that $n equiv 0 pmod n$.
EVERYTHING falls from that exceedingly nicely.
(In other words nearly all arithmetic distributes over modulus, and that $n$ is equivalent to $0$.)
(So $(n-1)^2$ is over modulus then, under $mod n$ arithmetic, the same thing as $(-1)^2$.)
(A few things to watch out for. Division doesn't distribute unless divisor and modulus are relatively prime. And the powers of exponents are not actually arithmetic so they don't distribute.)
edited Nov 20 '18 at 21:27
zwim
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
edited Nov 20 '18 at 21:27
zwim
11.5k729
edited Nov 20 '18 at 21:27
zwim
11.5k729
edited Nov 20 '18 at 21:27
zwim
11.5k729
11.5k729
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
answered Nov 20 '18 at 21:14
fleablood
68.3k22685
68.3k22685
add a comment |
add a comment |
Since $${n^2}equiv 0pmod n$$ $$-2nequiv 0pmod n$$
You have $${(n-1)^2}equiv ({n^2}-2n+1)equiv 0+0+1equiv1 pmod n$$
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
add a comment |
Since $${n^2}equiv 0pmod n$$ $$-2nequiv 0pmod n$$
You have $${(n-1)^2}equiv ({n^2}-2n+1)equiv 0+0+1equiv1 pmod n$$
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
add a comment |
Since $${n^2}equiv 0pmod n$$ $$-2nequiv 0pmod n$$
You have $${(n-1)^2}equiv ({n^2}-2n+1)equiv 0+0+1equiv1 pmod n$$
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
Since $${n^2}equiv 0pmod n$$ $$-2nequiv 0pmod n$$
You have $${(n-1)^2}equiv ({n^2}-2n+1)equiv 0+0+1equiv1 pmod n$$
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
answered Nov 20 '18 at 21:15
Dr. Mathva
919316
919316
add a comment |
add a comment |
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