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Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

Fixed a slip in Pomerance's slides
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Kevin Buzzard
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Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

added 43 characters in body
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Kim Morrison
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Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

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