Primes congruent to a mod n.
$begingroup$
Is there at least one prime p that is congruent to a mod n, where n can be any positive integer and a can be any non-negative integer less than n?
elementary-number-theory prime-numbers modular-arithmetic distribution-of-primes
asked Jan 18 at 1:55
KroneckerKronecker
1025
$endgroup$
add a comment |
$begingroup$
Is there at least one prime p that is congruent to a mod n, where n can be any positive integer and a can be any non-negative integer less than n?
elementary-number-theory prime-numbers modular-arithmetic distribution-of-primes
asked Jan 18 at 1:55
KroneckerKronecker
1025
$endgroup$
3
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
2
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
1
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49
add a comment |
$begingroup$
Is there at least one prime p that is congruent to a mod n, where n can be any positive integer and a can be any non-negative integer less than n?
elementary-number-theory prime-numbers modular-arithmetic distribution-of-primes
asked Jan 18 at 1:55
KroneckerKronecker
1025
$endgroup$
Is there at least one prime p that is congruent to a mod n, where n can be any positive integer and a can be any non-negative integer less than n?
elementary-number-theory prime-numbers modular-arithmetic distribution-of-primes
elementary-number-theory prime-numbers modular-arithmetic distribution-of-primes
asked Jan 18 at 1:55
KroneckerKronecker
1025
asked Jan 18 at 1:55
KroneckerKronecker
1025
edited Jan 18 at 17:13
Kronecker
asked Jan 18 at 1:55
KroneckerKronecker
1025
asked Jan 18 at 1:55
KroneckerKronecker
1025
asked Jan 18 at 1:55
KroneckerKronecker
1025
1025
3
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
2
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
1
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49
add a comment |
3
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
2
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
1
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49
3
3
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
2
2
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
1
1
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The complete answer is given by the different comments:
If $(a,n)=1$ then there are infinite primes that satisfy the equation $p=a$ mod $n$. This is the content of the Dirichlet's theorem on arithmetic progressions. Moreover, the primes are evenly distributed (asymptotically) among the congruence classes modulo $n$ containing the $a$'s coprime to $n$.
If $(a,n)>1$, we can have to cases:
1) If $a$ is prime, then we only have the trivial solution $p=a$.
Proof: If $a$ is prime and $(a,n)>1$ then $a$ is a factor of $n$, so $n=ab$ for some positive integer $b$, so the equation can be written $p = a$ mod $ab$, meaning $p$ = $a + (ab)c$ for some integer $c$. Then $p = a(1+bc)$, but $a$ is prime, so there is only one solution, namely when $c=0$.
2) If $a$ is composite, then there are no solutions.
Proof: $p = a$ mod $n$, means $p$ = $a + nI$ for some integer $I$, let's define $J=nI$ so now $p$ = $a + J$. Now, since $a$ and $n$ are not coprime, the same applies for $a$ and $J$, so by Bézout's identity: $aA+JB=C$ where $C > 1$ is the gcd of $a$ and $J$, and $A$ and $B$ are non-zero integers. Then $p$ = $a + ((C-aA)/B)$. And since $C$ is a factor of $a$ we have $p=CD + ((C-CDA)/B)$ for some positive integer $D$. Then $p=C(1+((1-DA)/B)$. Now, because $C>1$, for $p$ to be prime we need $(1+((1-DA)/B)$ to be equal to $1$, so $(1-DA)$ needs to be zero. But we said that $D$ is a positive integer, so $A$ has to be zero. But we said that $A$ is non-zero integer, we can conclude then that $p$ cannot be prime.
answered Jan 18 at 16:44
KroneckerKronecker
1025
$endgroup$
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The complete answer is given by the different comments:
If $(a,n)=1$ then there are infinite primes that satisfy the equation $p=a$ mod $n$. This is the content of the Dirichlet's theorem on arithmetic progressions. Moreover, the primes are evenly distributed (asymptotically) among the congruence classes modulo $n$ containing the $a$'s coprime to $n$.
If $(a,n)>1$, we can have to cases:
1) If $a$ is prime, then we only have the trivial solution $p=a$.
Proof: If $a$ is prime and $(a,n)>1$ then $a$ is a factor of $n$, so $n=ab$ for some positive integer $b$, so the equation can be written $p = a$ mod $ab$, meaning $p$ = $a + (ab)c$ for some integer $c$. Then $p = a(1+bc)$, but $a$ is prime, so there is only one solution, namely when $c=0$.
2) If $a$ is composite, then there are no solutions.
Proof: $p = a$ mod $n$, means $p$ = $a + nI$ for some integer $I$, let's define $J=nI$ so now $p$ = $a + J$. Now, since $a$ and $n$ are not coprime, the same applies for $a$ and $J$, so by Bézout's identity: $aA+JB=C$ where $C > 1$ is the gcd of $a$ and $J$, and $A$ and $B$ are non-zero integers. Then $p$ = $a + ((C-aA)/B)$. And since $C$ is a factor of $a$ we have $p=CD + ((C-CDA)/B)$ for some positive integer $D$. Then $p=C(1+((1-DA)/B)$. Now, because $C>1$, for $p$ to be prime we need $(1+((1-DA)/B)$ to be equal to $1$, so $(1-DA)$ needs to be zero. But we said that $D$ is a positive integer, so $A$ has to be zero. But we said that $A$ is non-zero integer, we can conclude then that $p$ cannot be prime.
answered Jan 18 at 16:44
KroneckerKronecker
1025
$endgroup$
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
add a comment |
$begingroup$
The complete answer is given by the different comments:
If $(a,n)=1$ then there are infinite primes that satisfy the equation $p=a$ mod $n$. This is the content of the Dirichlet's theorem on arithmetic progressions. Moreover, the primes are evenly distributed (asymptotically) among the congruence classes modulo $n$ containing the $a$'s coprime to $n$.
If $(a,n)>1$, we can have to cases:
1) If $a$ is prime, then we only have the trivial solution $p=a$.
Proof: If $a$ is prime and $(a,n)>1$ then $a$ is a factor of $n$, so $n=ab$ for some positive integer $b$, so the equation can be written $p = a$ mod $ab$, meaning $p$ = $a + (ab)c$ for some integer $c$. Then $p = a(1+bc)$, but $a$ is prime, so there is only one solution, namely when $c=0$.
2) If $a$ is composite, then there are no solutions.
Proof: $p = a$ mod $n$, means $p$ = $a + nI$ for some integer $I$, let's define $J=nI$ so now $p$ = $a + J$. Now, since $a$ and $n$ are not coprime, the same applies for $a$ and $J$, so by Bézout's identity: $aA+JB=C$ where $C > 1$ is the gcd of $a$ and $J$, and $A$ and $B$ are non-zero integers. Then $p$ = $a + ((C-aA)/B)$. And since $C$ is a factor of $a$ we have $p=CD + ((C-CDA)/B)$ for some positive integer $D$. Then $p=C(1+((1-DA)/B)$. Now, because $C>1$, for $p$ to be prime we need $(1+((1-DA)/B)$ to be equal to $1$, so $(1-DA)$ needs to be zero. But we said that $D$ is a positive integer, so $A$ has to be zero. But we said that $A$ is non-zero integer, we can conclude then that $p$ cannot be prime.
answered Jan 18 at 16:44
KroneckerKronecker
1025
$endgroup$
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
add a comment |
$begingroup$
The complete answer is given by the different comments:
If $(a,n)=1$ then there are infinite primes that satisfy the equation $p=a$ mod $n$. This is the content of the Dirichlet's theorem on arithmetic progressions. Moreover, the primes are evenly distributed (asymptotically) among the congruence classes modulo $n$ containing the $a$'s coprime to $n$.
If $(a,n)>1$, we can have to cases:
1) If $a$ is prime, then we only have the trivial solution $p=a$.
Proof: If $a$ is prime and $(a,n)>1$ then $a$ is a factor of $n$, so $n=ab$ for some positive integer $b$, so the equation can be written $p = a$ mod $ab$, meaning $p$ = $a + (ab)c$ for some integer $c$. Then $p = a(1+bc)$, but $a$ is prime, so there is only one solution, namely when $c=0$.
2) If $a$ is composite, then there are no solutions.
Proof: $p = a$ mod $n$, means $p$ = $a + nI$ for some integer $I$, let's define $J=nI$ so now $p$ = $a + J$. Now, since $a$ and $n$ are not coprime, the same applies for $a$ and $J$, so by Bézout's identity: $aA+JB=C$ where $C > 1$ is the gcd of $a$ and $J$, and $A$ and $B$ are non-zero integers. Then $p$ = $a + ((C-aA)/B)$. And since $C$ is a factor of $a$ we have $p=CD + ((C-CDA)/B)$ for some positive integer $D$. Then $p=C(1+((1-DA)/B)$. Now, because $C>1$, for $p$ to be prime we need $(1+((1-DA)/B)$ to be equal to $1$, so $(1-DA)$ needs to be zero. But we said that $D$ is a positive integer, so $A$ has to be zero. But we said that $A$ is non-zero integer, we can conclude then that $p$ cannot be prime.
answered Jan 18 at 16:44
KroneckerKronecker
1025
$endgroup$
The complete answer is given by the different comments:
If $(a,n)=1$ then there are infinite primes that satisfy the equation $p=a$ mod $n$. This is the content of the Dirichlet's theorem on arithmetic progressions. Moreover, the primes are evenly distributed (asymptotically) among the congruence classes modulo $n$ containing the $a$'s coprime to $n$.
If $(a,n)>1$, we can have to cases:
1) If $a$ is prime, then we only have the trivial solution $p=a$.
Proof: If $a$ is prime and $(a,n)>1$ then $a$ is a factor of $n$, so $n=ab$ for some positive integer $b$, so the equation can be written $p = a$ mod $ab$, meaning $p$ = $a + (ab)c$ for some integer $c$. Then $p = a(1+bc)$, but $a$ is prime, so there is only one solution, namely when $c=0$.
2) If $a$ is composite, then there are no solutions.
Proof: $p = a$ mod $n$, means $p$ = $a + nI$ for some integer $I$, let's define $J=nI$ so now $p$ = $a + J$. Now, since $a$ and $n$ are not coprime, the same applies for $a$ and $J$, so by Bézout's identity: $aA+JB=C$ where $C > 1$ is the gcd of $a$ and $J$, and $A$ and $B$ are non-zero integers. Then $p$ = $a + ((C-aA)/B)$. And since $C$ is a factor of $a$ we have $p=CD + ((C-CDA)/B)$ for some positive integer $D$. Then $p=C(1+((1-DA)/B)$. Now, because $C>1$, for $p$ to be prime we need $(1+((1-DA)/B)$ to be equal to $1$, so $(1-DA)$ needs to be zero. But we said that $D$ is a positive integer, so $A$ has to be zero. But we said that $A$ is non-zero integer, we can conclude then that $p$ cannot be prime.
answered Jan 18 at 16:44
KroneckerKronecker
1025
edited Jan 18 at 16:56
answered Jan 18 at 16:44
KroneckerKronecker
1025
answered Jan 18 at 16:44
KroneckerKronecker
1025
answered Jan 18 at 16:44
KroneckerKronecker
1025
1025
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
add a comment |
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Just one note: this solution assumes $aleq n$; we have for example at $a=6$ and $n=4$ the prime $p=2equiv 6bmod 4$. Another way to put it: for any $a$ where $gcd(a,n)>1$ all values $mequiv abmod n$ so that $m>n$ are composite.
$endgroup$
– Carl Schildkraut
Jan 18 at 16:53
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
$begingroup$
Thank you, I always thought that the congruence equations assumed that $a≤n$. I will edit the question so that this is explicit.
$endgroup$
– Kronecker
Jan 18 at 17:01
add a comment |
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3
$begingroup$
well, if $gcd(a,n) = 1,$ yes.
$endgroup$
– Will Jagy
Jan 18 at 1:58
2
$begingroup$
See Dirichlet's Theorem on Primes in Arithmetic Progressions
$endgroup$
– lulu
Jan 18 at 2:00
1
$begingroup$
for example, there is no prime $p equiv 15 pmod {30}$
$endgroup$
– Will Jagy
Jan 18 at 2:00
$begingroup$
So is there a result that establishes in which cases of a and n there is no prime p that satisfies the congruence?
$endgroup$
– Kronecker
Jan 18 at 2:55
$begingroup$
@user343996 If $a=0$ , then there is no such prime, if $n$ is composite. If $ane 0$ and $gcd(a,n)ne 1$, then there is no such prime, if $a$ is composite.
$endgroup$
– Peter
Jan 18 at 8:49