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Conjectures for which there are strong heuristic arguments both for and against
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I find the conjecture that there are infinitely many Fermat primes to be very interesting, because there is a very reasonable-seeming and semi-quantitative heuristic argument for the conjecture, but also an equally reasonable-seeming heuristic argument against it.
Are there any other conjectures or theorems such that one can provide a short and easy-to-understand heuristic argument for the conjecture, and an equally plausible heuristic argument against it? Preferably this would be for a conjecture and heuristic arguments that are easy for a non-expert to understand, and heuristic arguments that are plausible/precise enough that someone who wasn't previously familiar with the problem would be quite confident that either argument was correct until they heard the other one. Ideally, the conjecture would still remain an open problem. (I'm not interested in a long, detailed proof that is incorrect because of some subtle logical error, of which there have of course been many examples.)
My question is somewhat related to this one.
soft-question big-list conjectures open-problem
asked Jul 18 at 2:02
tparker
1,703629
 |Â
show 1 more comment
up vote
7
down vote
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I find the conjecture that there are infinitely many Fermat primes to be very interesting, because there is a very reasonable-seeming and semi-quantitative heuristic argument for the conjecture, but also an equally reasonable-seeming heuristic argument against it.
Are there any other conjectures or theorems such that one can provide a short and easy-to-understand heuristic argument for the conjecture, and an equally plausible heuristic argument against it? Preferably this would be for a conjecture and heuristic arguments that are easy for a non-expert to understand, and heuristic arguments that are plausible/precise enough that someone who wasn't previously familiar with the problem would be quite confident that either argument was correct until they heard the other one. Ideally, the conjecture would still remain an open problem. (I'm not interested in a long, detailed proof that is incorrect because of some subtle logical error, of which there have of course been many examples.)
My question is somewhat related to this one.
soft-question big-list conjectures open-problem
asked Jul 18 at 2:02
tparker
1,703629
I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
2
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11
 |Â
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I find the conjecture that there are infinitely many Fermat primes to be very interesting, because there is a very reasonable-seeming and semi-quantitative heuristic argument for the conjecture, but also an equally reasonable-seeming heuristic argument against it.
Are there any other conjectures or theorems such that one can provide a short and easy-to-understand heuristic argument for the conjecture, and an equally plausible heuristic argument against it? Preferably this would be for a conjecture and heuristic arguments that are easy for a non-expert to understand, and heuristic arguments that are plausible/precise enough that someone who wasn't previously familiar with the problem would be quite confident that either argument was correct until they heard the other one. Ideally, the conjecture would still remain an open problem. (I'm not interested in a long, detailed proof that is incorrect because of some subtle logical error, of which there have of course been many examples.)
My question is somewhat related to this one.
soft-question big-list conjectures open-problem
asked Jul 18 at 2:02
tparker
1,703629
I find the conjecture that there are infinitely many Fermat primes to be very interesting, because there is a very reasonable-seeming and semi-quantitative heuristic argument for the conjecture, but also an equally reasonable-seeming heuristic argument against it.
Are there any other conjectures or theorems such that one can provide a short and easy-to-understand heuristic argument for the conjecture, and an equally plausible heuristic argument against it? Preferably this would be for a conjecture and heuristic arguments that are easy for a non-expert to understand, and heuristic arguments that are plausible/precise enough that someone who wasn't previously familiar with the problem would be quite confident that either argument was correct until they heard the other one. Ideally, the conjecture would still remain an open problem. (I'm not interested in a long, detailed proof that is incorrect because of some subtle logical error, of which there have of course been many examples.)
My question is somewhat related to this one.
soft-question big-list conjectures open-problem
asked Jul 18 at 2:02
tparker
1,703629
asked Jul 18 at 2:02
tparker
1,703629
asked Jul 18 at 2:02
tparker
1,703629
asked Jul 18 at 2:02
tparker
1,703629
1,703629
I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
2
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11
 |Â
show 1 more comment
I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
2
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11
I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
2
2
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11
 |Â
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I find heuristic proofs to be rather sad. A proof or an argument should be written well enough so that it can be read by many. But...I do find your question interesting and noteworthy, +1!
– Prime
Jul 18 at 3:56
2
+1, I like the question -- but I don't agree that the arguments for and against infinitely many Fermat primes are "equally reasonable". In the Wikipedia section you link to, the first two arguments use less information we have about the Fermat prime candidates than the third one. I can argue that square numbers are more likely to be prime because I know they must contain even numbers of each prime factor, so the probability of being divisible by prime $p$ is only $frac1p^2$. I just haven't used the information that square numbers are in fact composite.
– joriki
Jul 18 at 4:45
To have roughly equally reasonable arguments in a case like this (infinitely or finitely many primes of a certain form), you'd have to have two pieces of information that go in opposite directions, one making prime factors more likely and one less. Then it wouldn't be immediately clear what would result from combining all the information.
– joriki
Jul 18 at 4:57
In the case of the Fermat primes, though, we have one calculation without information that predicts finitely many Fermat primes, one calculation with information $A$ that suggests that prime factors are less likely that predicts more Fermat primes but still finitely many, and one calculation with information $B$ that also suggests that prime factors are less likely but is stronger and thus predicts infinitely many Fermat primes. The combination $Aland B$ would then predict infinitely many Fermat primes; there's no symmetry.
– joriki
Jul 18 at 4:57
I'm not arguing against your assessment that someone who heard either argument without the other might be convinced either way; I'm only arguing against your assessment that they're "equally reasonable-seeming". I accidentally misquoted that as "equally reasonable" above. If the "-seeming" part is meant to be evaluated in the absence of the other argument, then you might be right. I'm just saying that if you do have both arguments, then they neither are, nor seem equally reasonable.
– joriki
Jul 18 at 5:11