When for given $p$ and $n$ we have $p^n pmod{4}=1$?
Consider $p$ is a prime odd number and $n$ is an integer positive number.
My question: Is there a condition over parameters $p$ and $n$ such that
$p^n equiv1pmod{4}$ ?
Thanks for any suggestions.
elementary-number-theory
edited Dec 1 at 17:19
Yadati Kiran
1,702519
asked Dec 1 at 17:14
user0410
1898
add a comment |
Consider $p$ is a prime odd number and $n$ is an integer positive number.
My question: Is there a condition over parameters $p$ and $n$ such that
$p^n equiv1pmod{4}$ ?
Thanks for any suggestions.
elementary-number-theory
edited Dec 1 at 17:19
Yadati Kiran
1,702519
asked Dec 1 at 17:14
user0410
1898
add a comment |
Consider $p$ is a prime odd number and $n$ is an integer positive number.
My question: Is there a condition over parameters $p$ and $n$ such that
$p^n equiv1pmod{4}$ ?
Thanks for any suggestions.
elementary-number-theory
edited Dec 1 at 17:19
Yadati Kiran
1,702519
asked Dec 1 at 17:14
user0410
1898
Consider $p$ is a prime odd number and $n$ is an integer positive number.
My question: Is there a condition over parameters $p$ and $n$ such that
$p^n equiv1pmod{4}$ ?
Thanks for any suggestions.
elementary-number-theory
elementary-number-theory
edited Dec 1 at 17:19
Yadati Kiran
1,702519
asked Dec 1 at 17:14
user0410
1898
edited Dec 1 at 17:19
Yadati Kiran
1,702519
asked Dec 1 at 17:14
user0410
1898
edited Dec 1 at 17:19
Yadati Kiran
1,702519
edited Dec 1 at 17:19
Yadati Kiran
1,702519
edited Dec 1 at 17:19
Yadati Kiran
1,702519
1,702519
asked Dec 1 at 17:14
user0410
1898
asked Dec 1 at 17:14
user0410
1898
asked Dec 1 at 17:14
user0410
1898
1898
add a comment |
add a comment |
1 Answer
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Since $p$ is odd, either $p = 4k+1$ or $p=4k-1$ for some $k$. Applying the binomial theorem, we see that in the former case,
$$ p^n = (4k+1)^n = 1+ sum_{j=1}^n binom{n}{j} (4k)^j equiv 1 , (text{mod } 4), $$
so the required condition holds for all such primes.
In the latter case,
$$ p^n = (4k-1)^n = (-1)^n + sum_{j=1}^n binom{n}{j} (4k)^j (-1)^{n-j} equiv (-1)^n , (text{mod } 4), $$
so $n$ should be even.
answered Dec 1 at 17:19
MisterRiemann
5,7691624
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
add a comment |
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Since $p$ is odd, either $p = 4k+1$ or $p=4k-1$ for some $k$. Applying the binomial theorem, we see that in the former case,
$$ p^n = (4k+1)^n = 1+ sum_{j=1}^n binom{n}{j} (4k)^j equiv 1 , (text{mod } 4), $$
so the required condition holds for all such primes.
In the latter case,
$$ p^n = (4k-1)^n = (-1)^n + sum_{j=1}^n binom{n}{j} (4k)^j (-1)^{n-j} equiv (-1)^n , (text{mod } 4), $$
so $n$ should be even.
answered Dec 1 at 17:19
MisterRiemann
5,7691624
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
add a comment |
Since $p$ is odd, either $p = 4k+1$ or $p=4k-1$ for some $k$. Applying the binomial theorem, we see that in the former case,
$$ p^n = (4k+1)^n = 1+ sum_{j=1}^n binom{n}{j} (4k)^j equiv 1 , (text{mod } 4), $$
so the required condition holds for all such primes.
In the latter case,
$$ p^n = (4k-1)^n = (-1)^n + sum_{j=1}^n binom{n}{j} (4k)^j (-1)^{n-j} equiv (-1)^n , (text{mod } 4), $$
so $n$ should be even.
answered Dec 1 at 17:19
MisterRiemann
5,7691624
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
add a comment |
Since $p$ is odd, either $p = 4k+1$ or $p=4k-1$ for some $k$. Applying the binomial theorem, we see that in the former case,
$$ p^n = (4k+1)^n = 1+ sum_{j=1}^n binom{n}{j} (4k)^j equiv 1 , (text{mod } 4), $$
so the required condition holds for all such primes.
In the latter case,
$$ p^n = (4k-1)^n = (-1)^n + sum_{j=1}^n binom{n}{j} (4k)^j (-1)^{n-j} equiv (-1)^n , (text{mod } 4), $$
so $n$ should be even.
answered Dec 1 at 17:19
MisterRiemann
5,7691624
Since $p$ is odd, either $p = 4k+1$ or $p=4k-1$ for some $k$. Applying the binomial theorem, we see that in the former case,
$$ p^n = (4k+1)^n = 1+ sum_{j=1}^n binom{n}{j} (4k)^j equiv 1 , (text{mod } 4), $$
so the required condition holds for all such primes.
In the latter case,
$$ p^n = (4k-1)^n = (-1)^n + sum_{j=1}^n binom{n}{j} (4k)^j (-1)^{n-j} equiv (-1)^n , (text{mod } 4), $$
so $n$ should be even.
answered Dec 1 at 17:19
MisterRiemann
5,7691624
answered Dec 1 at 17:19
MisterRiemann
5,7691624
answered Dec 1 at 17:19
MisterRiemann
5,7691624
answered Dec 1 at 17:19
MisterRiemann
5,7691624
5,7691624
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
add a comment |
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
I would very much appreciate for your nice solution.
– user0410
Dec 1 at 17:27
1
1
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
@user0410 Glad that it helped!
– MisterRiemann
Dec 1 at 17:29
2
2
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
If you reduce the prime $mod{4}$ prior to exponentiation, you can avoid the need for the binomial expansion. $pequiv pm1 mod{4}$, so $p^nequiv (pm1)^nmod{4}$. Keeping proper track of the $pm$ signs, you get the result given by MisterRiemann.
– Keith Backman
Dec 2 at 3:09
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
@KeithBackman Good point, not really sure how I managed to miss that!
– MisterRiemann
Dec 2 at 9:27
add a comment |
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