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Lagrange four squares theorem
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Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$.
$bullet$ Is $X=0,1,2,ldots$?
$bullet$ If not what is a minimal set $X$ with the given property?
nt.number-theory algebraic-number-theory additive-combinatorics
asked 5 hours ago
M. Farrokhi D. G.
840411
add a comment |Â
up vote
7
down vote
favorite
Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$.
$bullet$ Is $X=0,1,2,ldots$?
$bullet$ If not what is a minimal set $X$ with the given property?
nt.number-theory algebraic-number-theory additive-combinatorics
asked 5 hours ago
M. Farrokhi D. G.
840411
1
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$.
$bullet$ Is $X=0,1,2,ldots$?
$bullet$ If not what is a minimal set $X$ with the given property?
nt.number-theory algebraic-number-theory additive-combinatorics
asked 5 hours ago
M. Farrokhi D. G.
840411
Lagrange's four square theorem states that every non-negative integer is a sum of squares of four non-negative integers. Suppose $X$ is a subset of non-negative integers with the same property, that is, every non-negative integer is a sum of squares of four elements of $X$.
$bullet$ Is $X=0,1,2,ldots$?
$bullet$ If not what is a minimal set $X$ with the given property?
nt.number-theory algebraic-number-theory additive-combinatorics
nt.number-theory algebraic-number-theory additive-combinatorics
asked 5 hours ago
M. Farrokhi D. G.
840411
asked 5 hours ago
M. Farrokhi D. G.
840411
edited 1 hour ago
asked 5 hours ago
M. Farrokhi D. G.
840411
asked 5 hours ago
M. Farrokhi D. G.
840411
asked 5 hours ago
M. Farrokhi D. G.
840411
840411
1
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago
add a comment |Â
1
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago
1
1
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago
add a comment |Â
1 Answer
1
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oldest
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up vote
3
down vote
The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $ 0, 1, 2, ldots setminus S $ for an infinite $S$.
For any $varepsilon>0$, Erdös and Nathanson proved the existence of a set $X$ with $|X cap [0,n]| = O(n^frac34 +varepsilon)$, so that already provides an upper bound for your second question.
The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X cap [0,n]| = O(n^1/2log^1/2 n)$. As the lower bound $|X cap [0,n]| =Omega(n^1/2)$ is obvious, this leaves a very small gap for improvement.
Spencer has found a different proof of Wirsing's result.
Other relevant references may be found in the second page of a paper of Vu. Note that most of these proofs are probabilistic.
answered 59 mins ago
Ofir Gorodetsky
4,92211934
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $ 0, 1, 2, ldots setminus S $ for an infinite $S$.
For any $varepsilon>0$, Erdös and Nathanson proved the existence of a set $X$ with $|X cap [0,n]| = O(n^frac34 +varepsilon)$, so that already provides an upper bound for your second question.
The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X cap [0,n]| = O(n^1/2log^1/2 n)$. As the lower bound $|X cap [0,n]| =Omega(n^1/2)$ is obvious, this leaves a very small gap for improvement.
Spencer has found a different proof of Wirsing's result.
Other relevant references may be found in the second page of a paper of Vu. Note that most of these proofs are probabilistic.
answered 59 mins ago
Ofir Gorodetsky
4,92211934
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
add a comment |Â
up vote
3
down vote
The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $ 0, 1, 2, ldots setminus S $ for an infinite $S$.
For any $varepsilon>0$, Erdös and Nathanson proved the existence of a set $X$ with $|X cap [0,n]| = O(n^frac34 +varepsilon)$, so that already provides an upper bound for your second question.
The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X cap [0,n]| = O(n^1/2log^1/2 n)$. As the lower bound $|X cap [0,n]| =Omega(n^1/2)$ is obvious, this leaves a very small gap for improvement.
Spencer has found a different proof of Wirsing's result.
Other relevant references may be found in the second page of a paper of Vu. Note that most of these proofs are probabilistic.
answered 59 mins ago
Ofir Gorodetsky
4,92211934
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $ 0, 1, 2, ldots setminus S $ for an infinite $S$.
For any $varepsilon>0$, Erdös and Nathanson proved the existence of a set $X$ with $|X cap [0,n]| = O(n^frac34 +varepsilon)$, so that already provides an upper bound for your second question.
The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X cap [0,n]| = O(n^1/2log^1/2 n)$. As the lower bound $|X cap [0,n]| =Omega(n^1/2)$ is obvious, this leaves a very small gap for improvement.
Spencer has found a different proof of Wirsing's result.
Other relevant references may be found in the second page of a paper of Vu. Note that most of these proofs are probabilistic.
answered 59 mins ago
Ofir Gorodetsky
4,92211934
The set $X$ doesn't have to be the set of non-negative integers. This was known already to Härtter and Zöllner in 1977, who constructed an $X$ of the form $ 0, 1, 2, ldots setminus S $ for an infinite $S$.
For any $varepsilon>0$, Erdös and Nathanson proved the existence of a set $X$ with $|X cap [0,n]| = O(n^frac34 +varepsilon)$, so that already provides an upper bound for your second question.
The problem was essentially settled by Wirsing in 1986, who proved that one has $X$ with $|X cap [0,n]| = O(n^1/2log^1/2 n)$. As the lower bound $|X cap [0,n]| =Omega(n^1/2)$ is obvious, this leaves a very small gap for improvement.
Spencer has found a different proof of Wirsing's result.
Other relevant references may be found in the second page of a paper of Vu. Note that most of these proofs are probabilistic.
answered 59 mins ago
Ofir Gorodetsky
4,92211934
edited 29 mins ago
answered 59 mins ago
Ofir Gorodetsky
4,92211934
answered 59 mins ago
Ofir Gorodetsky
4,92211934
answered 59 mins ago
Ofir Gorodetsky
4,92211934
4,92211934
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
add a comment |Â
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
1
1
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
I guess $0, 1, 2, dotsc/S$ should be $0, 1, 2, dotsc setminus S$, right?
– LSpice
39 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
@LSpice Yes, thank you. Corrected now.
– Ofir Gorodetsky
36 mins ago
add a comment |Â
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1
Sieve theory will give that one can take $X$ to be the set of numbers with at most $r$ prime factors for some enormous $r$, probably
– Stanley Yao Xiao
4 hours ago