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Is every positive integer greater than $2$ the sum of a prime and two squares?
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I'm not sure if this conjecture is less hard than Goldbachs conjecture:
any integer greater than $2$ is the sum of an odd prime and two squares of integers.
Facts as:
Every prime of the form $4n+1$ is the sum of two squares.
Every natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $10^6$.
Even if the conjecture maybe is to hard to prove I would like to see ideas and heuristics about it.
Or counter-examples!
number-theory prime-numbers goldbachs-conjecture
edited Aug 6 at 11:08
Peter
45.1k939119
asked Aug 5 at 13:12
Lehs
6,82431561
 |Â
show 7 more comments
up vote
3
down vote
favorite
2
I'm not sure if this conjecture is less hard than Goldbachs conjecture:
any integer greater than $2$ is the sum of an odd prime and two squares of integers.
Facts as:
Every prime of the form $4n+1$ is the sum of two squares.
Every natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $10^6$.
Even if the conjecture maybe is to hard to prove I would like to see ideas and heuristics about it.
Or counter-examples!
number-theory prime-numbers goldbachs-conjecture
edited Aug 6 at 11:08
Peter
45.1k939119
asked Aug 5 at 13:12
Lehs
6,82431561
There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
2
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
1
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
1
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30
 |Â
show 7 more comments
up vote
3
down vote
favorite
2
up vote
3
down vote
favorite
2
2
I'm not sure if this conjecture is less hard than Goldbachs conjecture:
any integer greater than $2$ is the sum of an odd prime and two squares of integers.
Facts as:
Every prime of the form $4n+1$ is the sum of two squares.
Every natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $10^6$.
Even if the conjecture maybe is to hard to prove I would like to see ideas and heuristics about it.
Or counter-examples!
number-theory prime-numbers goldbachs-conjecture
edited Aug 6 at 11:08
Peter
45.1k939119
asked Aug 5 at 13:12
Lehs
6,82431561
I'm not sure if this conjecture is less hard than Goldbachs conjecture:
any integer greater than $2$ is the sum of an odd prime and two squares of integers.
Facts as:
Every prime of the form $4n+1$ is the sum of two squares.
Every natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $10^6$.
Even if the conjecture maybe is to hard to prove I would like to see ideas and heuristics about it.
Or counter-examples!
number-theory prime-numbers goldbachs-conjecture
edited Aug 6 at 11:08
Peter
45.1k939119
asked Aug 5 at 13:12
Lehs
6,82431561
edited Aug 6 at 11:08
Peter
45.1k939119
edited Aug 6 at 11:08
Peter
45.1k939119
edited Aug 6 at 11:08
Peter
45.1k939119
45.1k939119
asked Aug 5 at 13:12
Lehs
6,82431561
asked Aug 5 at 13:12
Lehs
6,82431561
asked Aug 5 at 13:12
Lehs
6,82431561
6,82431561
There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
2
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
1
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
1
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30
 |Â
show 7 more comments
There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
2
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
1
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
1
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30
There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
2
2
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
1
1
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
1
1
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30
 |Â
show 7 more comments
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There where two conjectures and the remaining is: $n=p+a^2+b^2$.
– Lehs
Aug 5 at 14:46
"Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true?
– Keith Backman
Aug 5 at 17:33
2
By the way, $n$ can be expressed as a sum of two squares iff all primes $pmid n$ of form $p=4k+3$ divide $n$ in even power.
– Sil
Aug 5 at 17:52
1
@KeithBackman If I understand what you are asking, that is not true for four squares, because $2^2r+1$ can be expressed in exactly one way $2^2r+1=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares).
– Sil
Aug 5 at 20:10
1
@Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $nge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $nle 10^8$ and heuristically , it should always be true. All we need is a prime $ple n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,cdots$. Maybe , someone can work that out ...
– Peter
Aug 7 at 16:30