Special cases of Szemeredi's Theorem?
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked yesterday
Isky Mathews
878214
add a comment |
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked yesterday
Isky Mathews
878214
add a comment |
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked yesterday
Isky Mathews
878214
Are there examples of special cases of Szemeredi's theorem which one can give, which are slightly non-trivial?
To clarify, I'm looking for sets of integers where we can show that they contain infinitely many arithmetic progressions of any finite length. When I say "slightly non-trivial", this is a non-well-defined condition which attempts to remove commenters saying things like "the even numbers" or "all infinite arithmetic progressions" - I'm looking for something like a proof for the special case of the square-free integers.
(Final comment: This is less of a question that I particularly need answered but more of an attempt to collate a collection of such "slightly non-trivial" proofs for my and others' appreciation.)
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
combinatorics number-theory soft-question alternative-proof arithmetic-progressions
asked yesterday
Isky Mathews
878214
asked yesterday
Isky Mathews
878214
asked yesterday
Isky Mathews
878214
asked yesterday
Isky Mathews
878214
asked yesterday
Isky Mathews
878214
878214
add a comment |
add a comment |
1 Answer
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oldest
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The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered yesterday
Sandeep Silwal
5,84311236
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered yesterday
Sandeep Silwal
5,84311236
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
add a comment |
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered yesterday
Sandeep Silwal
5,84311236
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
add a comment |
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered yesterday
Sandeep Silwal
5,84311236
The Green-Tao theorem comes to mind. It states that the primes contain arithmetic progressions of arbitrary length. If you want to search for other 'non-trivial' special cases, starting with this Erdos conjecture might be a good idea.
There are also really interesting special cases of Szemeredi's Theorem that were proved before and are more accessible such as Roth's Theorem. This theorem states that any subset of integers of positive density contains a $3$ term arithmetic progression. This theorem is particularly interesting because there has been work on improving the density value. A good survey article is here.
answered yesterday
Sandeep Silwal
5,84311236
answered yesterday
Sandeep Silwal
5,84311236
answered yesterday
Sandeep Silwal
5,84311236
answered yesterday
Sandeep Silwal
5,84311236
5,84311236
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
add a comment |
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
While I thank you for the time taken to write your answer, what I meant by slightly non-trivial is that the proofs in question are to be short enough to be contained in a MSE post, assuming potentially certain standard NT theorems (like PNT or Stirling's Approximation). Green-Tao's proof is extremely complex and furthermore assumes Szemeredi's Theorem as part of the proof, so definitely isn't a special case.
– Isky Mathews
yesterday
add a comment |
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