$D := (-1)^{frac{m-1}{2}}n$ and $N := (-1)^{frac{n-1}{2}}n$. Show that $N = D$.
$begingroup$
Let $m,n$ be an odd, positive integer:
$D := (-1)^{frac{m-1}{2}}n$ with D $equiv 1$ (mod 4)
$N := (-1)^{frac{n-1}{2}}n$ with N $equiv 1$ (mod 4)
Show that $N = D$:
What I got so far:
$D = pm n$ and it has to be odd since $n$ is odd. Same for $N$. Why can I follow from that that $D = N$?
linear-algebra abstract-algebra number-theory modular-arithmetic
edited Jan 22 at 14:03
Yanko
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
$endgroup$
add a comment |
$begingroup$
Let $m,n$ be an odd, positive integer:
$D := (-1)^{frac{m-1}{2}}n$ with D $equiv 1$ (mod 4)
$N := (-1)^{frac{n-1}{2}}n$ with N $equiv 1$ (mod 4)
Show that $N = D$:
What I got so far:
$D = pm n$ and it has to be odd since $n$ is odd. Same for $N$. Why can I follow from that that $D = N$?
linear-algebra abstract-algebra number-theory modular-arithmetic
edited Jan 22 at 14:03
Yanko
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
$endgroup$
add a comment |
$begingroup$
Let $m,n$ be an odd, positive integer:
$D := (-1)^{frac{m-1}{2}}n$ with D $equiv 1$ (mod 4)
$N := (-1)^{frac{n-1}{2}}n$ with N $equiv 1$ (mod 4)
Show that $N = D$:
What I got so far:
$D = pm n$ and it has to be odd since $n$ is odd. Same for $N$. Why can I follow from that that $D = N$?
linear-algebra abstract-algebra number-theory modular-arithmetic
edited Jan 22 at 14:03
Yanko
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
$endgroup$
Let $m,n$ be an odd, positive integer:
$D := (-1)^{frac{m-1}{2}}n$ with D $equiv 1$ (mod 4)
$N := (-1)^{frac{n-1}{2}}n$ with N $equiv 1$ (mod 4)
Show that $N = D$:
What I got so far:
$D = pm n$ and it has to be odd since $n$ is odd. Same for $N$. Why can I follow from that that $D = N$?
linear-algebra abstract-algebra number-theory modular-arithmetic
linear-algebra abstract-algebra number-theory modular-arithmetic
edited Jan 22 at 14:03
Yanko
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
edited Jan 22 at 14:03
Yanko
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
edited Jan 22 at 14:03
Yanko
7,2201729
edited Jan 22 at 14:03
Yanko
7,2201729
edited Jan 22 at 14:03
Yanko
7,2201729
7,2201729
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
asked Jan 22 at 14:01
FabianSchneiderFabianSchneider
706
706
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
From the assumption each of $N,D$ is either $n$ or $-n$.
If $n equiv 1 text{ mod } 4$ then $-n equiv 3 text{ mod } 4$ and vice versa.
Is it possible that $N=n$ and $D=-n$?
answered Jan 22 at 14:05
YankoYanko
7,2201729
$endgroup$
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
From the assumption each of $N,D$ is either $n$ or $-n$.
If $n equiv 1 text{ mod } 4$ then $-n equiv 3 text{ mod } 4$ and vice versa.
Is it possible that $N=n$ and $D=-n$?
answered Jan 22 at 14:05
YankoYanko
7,2201729
$endgroup$
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
add a comment |
$begingroup$
Hint:
From the assumption each of $N,D$ is either $n$ or $-n$.
If $n equiv 1 text{ mod } 4$ then $-n equiv 3 text{ mod } 4$ and vice versa.
Is it possible that $N=n$ and $D=-n$?
answered Jan 22 at 14:05
YankoYanko
7,2201729
$endgroup$
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
add a comment |
$begingroup$
Hint:
From the assumption each of $N,D$ is either $n$ or $-n$.
If $n equiv 1 text{ mod } 4$ then $-n equiv 3 text{ mod } 4$ and vice versa.
Is it possible that $N=n$ and $D=-n$?
answered Jan 22 at 14:05
YankoYanko
7,2201729
$endgroup$
Hint:
From the assumption each of $N,D$ is either $n$ or $-n$.
If $n equiv 1 text{ mod } 4$ then $-n equiv 3 text{ mod } 4$ and vice versa.
Is it possible that $N=n$ and $D=-n$?
answered Jan 22 at 14:05
YankoYanko
7,2201729
answered Jan 22 at 14:05
YankoYanko
7,2201729
answered Jan 22 at 14:05
YankoYanko
7,2201729
answered Jan 22 at 14:05
YankoYanko
7,2201729
7,2201729
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
add a comment |
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
So because $n equiv 1$ (mod 4) implies $-n equiv 3$ (mod 4), both $N$ and $D$ has to be $+n$. And $n equiv 3$ (mod 4) implies $-n equiv 1$ (mod 4), so both $N$ and $D$ has to be $-n$. And we are done because that are all cases for $n$. Is that what you telling me?
$endgroup$
– FabianSchneider
Jan 22 at 14:13
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
@FabianSchneider Exactly. Good job.
$endgroup$
– Yanko
Jan 22 at 14:18
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
$begingroup$
Perfect :) thanks a lot
$endgroup$
– FabianSchneider
Jan 22 at 14:20
add a comment |
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