Is there a 2-power-twinless prime?
$begingroup$
Call two primes 2-power-twins if their difference is (can you guess?) a power of 2.
For example, 11 and 19 are 2-power-twins.
Is there a 2-power-twinless prime?
I would imagine that this is doable the following way.
If I take a prime of the form $3k+1$, then I know that adding an odd power of 2 or subtracting an even power of 2 cannot give a prime.
If I take a prime of the form $5k+1$, then I know that adding a power of 2 that is $2bmod 4$ cannot give a prime.
Do such observations give enough conditions to conclude the existence of a 2-power-twinless prime from Dirichlet's theorem?
nt.number-theory prime-numbers
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
$endgroup$
add a comment |
$begingroup$
Call two primes 2-power-twins if their difference is (can you guess?) a power of 2.
For example, 11 and 19 are 2-power-twins.
Is there a 2-power-twinless prime?
I would imagine that this is doable the following way.
If I take a prime of the form $3k+1$, then I know that adding an odd power of 2 or subtracting an even power of 2 cannot give a prime.
If I take a prime of the form $5k+1$, then I know that adding a power of 2 that is $2bmod 4$ cannot give a prime.
Do such observations give enough conditions to conclude the existence of a 2-power-twinless prime from Dirichlet's theorem?
nt.number-theory prime-numbers
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
$endgroup$
2
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
2
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
2
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
add a comment |
$begingroup$
Call two primes 2-power-twins if their difference is (can you guess?) a power of 2.
For example, 11 and 19 are 2-power-twins.
Is there a 2-power-twinless prime?
I would imagine that this is doable the following way.
If I take a prime of the form $3k+1$, then I know that adding an odd power of 2 or subtracting an even power of 2 cannot give a prime.
If I take a prime of the form $5k+1$, then I know that adding a power of 2 that is $2bmod 4$ cannot give a prime.
Do such observations give enough conditions to conclude the existence of a 2-power-twinless prime from Dirichlet's theorem?
nt.number-theory prime-numbers
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
$endgroup$
Call two primes 2-power-twins if their difference is (can you guess?) a power of 2.
For example, 11 and 19 are 2-power-twins.
Is there a 2-power-twinless prime?
I would imagine that this is doable the following way.
If I take a prime of the form $3k+1$, then I know that adding an odd power of 2 or subtracting an even power of 2 cannot give a prime.
If I take a prime of the form $5k+1$, then I know that adding a power of 2 that is $2bmod 4$ cannot give a prime.
Do such observations give enough conditions to conclude the existence of a 2-power-twinless prime from Dirichlet's theorem?
nt.number-theory prime-numbers
nt.number-theory prime-numbers
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
asked Dec 4 '18 at 9:41
domotorpdomotorp
9,7473187
9,7473187
2
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
2
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
2
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
add a comment |
2
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
2
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
2
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
2
2
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
2
2
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
2
2
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Erdos proved that there is an arithmetic progression of odd numbers, none of which can be expressed as a sum of a power of two and a prime. I believe one can arrange for such an arithmetic progression to satisfy the hypotheses of Dirichlet's Theorem on primes in arithmetic progression, so that means there are infinitely many primes $p$ for which there is no prime $q$ such that $p-q$ is a power of two. So there are infinitely many primes $p$ that are not the larger of a pair of two-power-twins.
The Erdos result, from the 1950 paper in which he introduced covering congruences, has been expanded upon. I think it has been proved that there is an arithmetic progression of odd numbers $n$ such that $n$ is neither of the form $2^k+q$ nor of the form $q-2^k$ for any prime $q$ (but I don't have easy access to a citation for this). Modulo satisfying the Dirichlet hypothesis, this would establish the existence of infinitely many 2-power-twinless primes.
I think the appropriate citation is F. Cohen, J.L. Selfridge, Not every number is the sum or difference of two prime powers, Math. Comp. 29 (1975) 79–81. Theorem 1 states, there exists an arithmetic progression of odd numbers which are neither the sum nor difference of a power of two and a prime. They give the example, 47867742232066880047611079 is prime and neither the sum nor difference of a power of two and a prime.
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
$endgroup$
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Erdos proved that there is an arithmetic progression of odd numbers, none of which can be expressed as a sum of a power of two and a prime. I believe one can arrange for such an arithmetic progression to satisfy the hypotheses of Dirichlet's Theorem on primes in arithmetic progression, so that means there are infinitely many primes $p$ for which there is no prime $q$ such that $p-q$ is a power of two. So there are infinitely many primes $p$ that are not the larger of a pair of two-power-twins.
The Erdos result, from the 1950 paper in which he introduced covering congruences, has been expanded upon. I think it has been proved that there is an arithmetic progression of odd numbers $n$ such that $n$ is neither of the form $2^k+q$ nor of the form $q-2^k$ for any prime $q$ (but I don't have easy access to a citation for this). Modulo satisfying the Dirichlet hypothesis, this would establish the existence of infinitely many 2-power-twinless primes.
I think the appropriate citation is F. Cohen, J.L. Selfridge, Not every number is the sum or difference of two prime powers, Math. Comp. 29 (1975) 79–81. Theorem 1 states, there exists an arithmetic progression of odd numbers which are neither the sum nor difference of a power of two and a prime. They give the example, 47867742232066880047611079 is prime and neither the sum nor difference of a power of two and a prime.
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
$endgroup$
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
add a comment |
$begingroup$
Erdos proved that there is an arithmetic progression of odd numbers, none of which can be expressed as a sum of a power of two and a prime. I believe one can arrange for such an arithmetic progression to satisfy the hypotheses of Dirichlet's Theorem on primes in arithmetic progression, so that means there are infinitely many primes $p$ for which there is no prime $q$ such that $p-q$ is a power of two. So there are infinitely many primes $p$ that are not the larger of a pair of two-power-twins.
The Erdos result, from the 1950 paper in which he introduced covering congruences, has been expanded upon. I think it has been proved that there is an arithmetic progression of odd numbers $n$ such that $n$ is neither of the form $2^k+q$ nor of the form $q-2^k$ for any prime $q$ (but I don't have easy access to a citation for this). Modulo satisfying the Dirichlet hypothesis, this would establish the existence of infinitely many 2-power-twinless primes.
I think the appropriate citation is F. Cohen, J.L. Selfridge, Not every number is the sum or difference of two prime powers, Math. Comp. 29 (1975) 79–81. Theorem 1 states, there exists an arithmetic progression of odd numbers which are neither the sum nor difference of a power of two and a prime. They give the example, 47867742232066880047611079 is prime and neither the sum nor difference of a power of two and a prime.
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
$endgroup$
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
add a comment |
$begingroup$
Erdos proved that there is an arithmetic progression of odd numbers, none of which can be expressed as a sum of a power of two and a prime. I believe one can arrange for such an arithmetic progression to satisfy the hypotheses of Dirichlet's Theorem on primes in arithmetic progression, so that means there are infinitely many primes $p$ for which there is no prime $q$ such that $p-q$ is a power of two. So there are infinitely many primes $p$ that are not the larger of a pair of two-power-twins.
The Erdos result, from the 1950 paper in which he introduced covering congruences, has been expanded upon. I think it has been proved that there is an arithmetic progression of odd numbers $n$ such that $n$ is neither of the form $2^k+q$ nor of the form $q-2^k$ for any prime $q$ (but I don't have easy access to a citation for this). Modulo satisfying the Dirichlet hypothesis, this would establish the existence of infinitely many 2-power-twinless primes.
I think the appropriate citation is F. Cohen, J.L. Selfridge, Not every number is the sum or difference of two prime powers, Math. Comp. 29 (1975) 79–81. Theorem 1 states, there exists an arithmetic progression of odd numbers which are neither the sum nor difference of a power of two and a prime. They give the example, 47867742232066880047611079 is prime and neither the sum nor difference of a power of two and a prime.
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
$endgroup$
Erdos proved that there is an arithmetic progression of odd numbers, none of which can be expressed as a sum of a power of two and a prime. I believe one can arrange for such an arithmetic progression to satisfy the hypotheses of Dirichlet's Theorem on primes in arithmetic progression, so that means there are infinitely many primes $p$ for which there is no prime $q$ such that $p-q$ is a power of two. So there are infinitely many primes $p$ that are not the larger of a pair of two-power-twins.
The Erdos result, from the 1950 paper in which he introduced covering congruences, has been expanded upon. I think it has been proved that there is an arithmetic progression of odd numbers $n$ such that $n$ is neither of the form $2^k+q$ nor of the form $q-2^k$ for any prime $q$ (but I don't have easy access to a citation for this). Modulo satisfying the Dirichlet hypothesis, this would establish the existence of infinitely many 2-power-twinless primes.
I think the appropriate citation is F. Cohen, J.L. Selfridge, Not every number is the sum or difference of two prime powers, Math. Comp. 29 (1975) 79–81. Theorem 1 states, there exists an arithmetic progression of odd numbers which are neither the sum nor difference of a power of two and a prime. They give the example, 47867742232066880047611079 is prime and neither the sum nor difference of a power of two and a prime.
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
edited Dec 4 '18 at 12:17
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
answered Dec 4 '18 at 11:57
Gerry MyersonGerry Myerson
30.4k5140182
30.4k5140182
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
add a comment |
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
1
1
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
$begingroup$
Thanks! In fact, the proof goes exactly the way I've sketched in my answer, just you needed 42 primes to finish it... Here is the full paper: ams.org/journals/mcom/1975-29-129/S0025-5718-1975-0376583-0/…
$endgroup$
– domotorp
Dec 4 '18 at 12:59
2
2
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
$begingroup$
See also my paper "On integers not of the form $pm p^apm q^b$ [Proc. Amer. Math. Soc., 128(2000), no.4, 997--1002] available from maths.nju.edu.cn/~zwsun/34p.pdf ..
$endgroup$
– Zhi-Wei Sun
Dec 4 '18 at 17:21
add a comment |
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2
$begingroup$
You should exclude the obvious fact that $2$ is a $2$-power-twinless prime.
$endgroup$
– Tony Huynh
Dec 4 '18 at 10:21
2
$begingroup$
Brun sieve gives a bound for $sum_{n le x} 1_{n in P, n-2^k in P}$
$endgroup$
– reuns
Dec 4 '18 at 10:28
2
$begingroup$
$ 2 + 2^0 = 3 $
$endgroup$
– Tom
Dec 4 '18 at 12:49
$begingroup$
@Tony See Tom's comment.
$endgroup$
– domotorp
Dec 4 '18 at 12:51
$begingroup$
@reuns I don't see why that's useful.
$endgroup$
– domotorp
Dec 4 '18 at 12:51