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edited Jan 24 2011 at 12:32 |
The question should ask I suspect that there are infinitely many exceptions however this would imply that that there are infinitely many Sophie Germain primes (2p+1 also prime). This is not known to be true although there is every reason to expect that there are about $\frac{n}{\log^2n}$ such less than $n$. $p=(k+1)(ab)+k(a+b)$ or equivalently is true exactly if $p\equiv (ab)\bmod ((a+b)+ab)$ with (k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $\mathbf{ab<p}$ Otherwise you are allowing (k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $k=0,a=1,b=p$.-1 \mod k+1$ The largest that $k$ could be is $\frac{p-2}{5}$ in case $(a,b)=(1,1)$. Of the 1229 course then $2p+1$ is not prime (if $p \ne 2$) It may be that there are only finitely many primes with $(k+1)p+k^2$ prime for all $k$ up to 10000, 190 are Sophie Germain primes $p,2p+1$ both p/5$ however we don't need them all prime. , just to not have any factor congruent to $k \mod k+1.$ The first few Sophie Germain primes are If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$. later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$
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edited Jan 21 2011 at 12:49 |
If $2p+1$ is composite there is a solution with $k=1$ and if it is prime then there is none with $k=1$. In general there is a solution with $k=K$ if and only if $(K+1)p+K^2$ factors as the product of two factors both congruent to -1 $\mod K$.
The question should ask about
$p=(k+1)(ab)+k(a+b)$ or equivalently $p\equiv (ab)\bmod ((a+b)+ab)$ with $\mathbf{ab<p}$
Otherwise you are allowing $k=0,a=1,b=p$.
Of the 1229 primes up to 10000, 190 are Sophie Germain primes $p,2p+1$ both prime. The first few are
$$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173, 179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above except 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$
In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. All of the exceptions are Sophie Germain primes. The largest Sophie Germain primes under 1000 are
$$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with the exceptions of 9371 and 9539.
If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.
later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$
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edited Jan 21 2011 at 8:06 |
The question should ask about
$p=(k+1)(ab)+k(a+b)$ or equivalently $p\equiv (ab)\bmod ((a+b)+ab)$ with $\mathbf{ab<p}$
Otherwise you are allowing $k=0,a=1,b=p$.
Of the 1229 primes up to 10000, 190 are Sophie Germain primes $p,2p+1$ both prime. The first few are
$$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173, 179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above except 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$
In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. All of the exceptions are Sophie Germain primes. The largest Sophie Germain primes under 1000 are
$$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with the exceptions of 9371 and 9539.
If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.
later $p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$
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edited Jan 21 2011 at 6:34 |
WAIT: In the "equivalent" form doesn't $a=1,b=p\ The question should ask about $select for p=(k+1)(ab)+k(a+b)$ or equivalently $x\equiv p\ \bmod p\equiv (2p+1)? ab)\bmod ((a+b)+ab)$ with $SO any integer can be written in this form. But I see that \mathbf{ab<p}$ Otherwise you want the additional condition are allowing $ab I suspect not but I can't prove it. k=0,a=1,b=p$.
Of the 1229 primes less than up to 10000, 190 are Sophie Germain primes $p,2p+1$ both prime. The first few are $$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173can not be expressed in this way, call them bad179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above except 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$ In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. The proportion All of bad the exceptions are Sophie Germain primesis slightly decreasing . The largest Sophie Germain primes under 1000 are $$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with some fluctuation but does not seem likely to get below 10%the exceptions of 9371 and 9539. If $p$ is NOT a Sophie Germain prime and $2p+1=(2a+1)(2b+1)$ then $p=2ab+a+b$.
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edited Jan 21 2011 at 5:11 |
WAIT: In the "equivalent" form doesn't $a=1,b=p\ $ select for $x\equiv p\ \bmod (2p+1)? $ SO any integer can be written in this form. But I see that you want the additional condition $ab
I suspect not but I can't prove it. Of the 1229 primes less than 10000, 173 can not be expressed in this way, call them bad. The proportion of bad primes is slightly decreasing with some fluctuation but does not seem likely to get below 10%.
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edited Jan 21 2011 at 4:46 |
Corrected WAIT: OK, I suspect so. Every doesn't $a=1,b=p\ $ select for $x\equiv p\ \bmod (2p+1)? $ SO any integer up to 10000 prime or not seems to can be of written in this formfor some $a,b$.
I suspect not but I can't prove it. Of the 1229 primes less than 10000, 173 can not be expressed in this way, call them bad. The proportion of bad primes is slightly decreasing with some fluctuation but does not seem likely to get below 10%.
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edited Jan 21 2011 at 4:40 |
Corrected: OK, If my calculations are correct (this time) then of the 1229 primes less than I suspect so. Every integer up to 10000 , these: $5003, 5039, 5081, 5303, 5333, 6101, 6173, 6323, 6449, 6581, 6761, 7349, 7433, 7883, 8693, 9371, 9539$ are prime or not seems to be of that this form . That is not in the OEIS. I will double check nowbut I did not want to leave the wrong answer up and don't know how to retract an answer. for some $a,b$
I suspect not but I can't prove it. Of the 1229 primes less than 10000, 173 can not be expressed in this way, call them bad. The proportion of bad primes is slightly decreasing with some fluctuation but does not seem likely to get below 10%.
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edited Jan 21 2011 at 4:31 |
Corrected: OK, If my calculations are correct (this time) then of the 1229 primes less than 10000, these: $5003, 5039, 5081, 5303, 5333, 6101, 6173, 6323, 6449, 6581, 6761, 7349, 7433, 7883, 8693, 9371, 9539$ are not of that form. That is not in the OEIS. I will double check nowbut I did not want to leave the wrong answer up and don't know how to retract an answer.
I suspect not but I can't prove it. Of the 1229 primes less than 10000, 173 can not be expressed in this way, call them bad. The proportion of bad primes is slightly decreasing with some fluctuation but does not seem likely to get below 10%.
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answered Jan 21 2011 at 4:10 |
I suspect not but I can't prove it. Of the 1229 primes less than 10000, 173 can not be expressed in this way, call them bad. The proportion of bad primes is slightly decreasing with some fluctuation but does not seem likely to get below 10%.
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