Mixing
Conjecture about primes and square sums
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
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:
any prime of type $4n+1$ is the sum of two squares
any natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $1000000$.
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
asked 18 hours ago
Lehs
6,77731459
 |Â
show 5 more comments
up vote
3
down vote
favorite
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:
any prime of type $4n+1$ is the sum of two squares
any natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $1000000$.
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
asked 18 hours ago
Lehs
6,77731459
1
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
1
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
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
14 hours ago
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
12 hours ago
 |Â
show 5 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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:
any prime of type $4n+1$ is the sum of two squares
any natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $1000000$.
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
asked 18 hours ago
Lehs
6,77731459
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:
any prime of type $4n+1$ is the sum of two squares
any natural number is the sum of four squares
may or may not be helpful.
I've tested the conjecture for all integers less than $1000000$.
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
asked 18 hours ago
Lehs
6,77731459
edited 17 hours ago
asked 18 hours ago
Lehs
6,77731459
asked 18 hours ago
Lehs
6,77731459
asked 18 hours ago
Lehs
6,77731459
6,77731459
1
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
1
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
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
14 hours ago
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
12 hours ago
 |Â
show 5 more comments
1
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
1
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
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
14 hours ago
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
12 hours ago
1
1
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
1
1
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
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
14 hours ago
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
14 hours ago
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
12 hours ago
@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
12 hours ago
 |Â
show 5 more comments
1 Answer
1
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Here are some relevant computations. Fix $N$ large. We are interested in $$int_0^1 P_N(alpha)S_N(alpha)^2e(-Nalpha)dalpha$$ where $$P_N(alpha) := sum_p le N e(palpha)$$ $$S_N(alpha) := sum_m le sqrtN e(m^2alpha)$$ $$e(beta) := e^2pi i beta.$$ If $alpha = fracab$ for $b le sqrtlog N$ (say), then $$P_N(alpha) = sum_p le N e(pfracab) = sum_j=0^b-1 sum_substackp le N \ pa equiv j pmodb e(fracjb) approx sum_j=0^b-1 e(fracjb) fracpi(N)phi(b) = fracmu(b)phi(b)pi(N).$$
And, $S_N(fracab) approx fracsqrtNbsum_m le b e(m^2fracab)$. By exact values for Guass sums, the square of the sum is $0$ if $b equiv 2 pmod4$; $b$ if $b equiv 1 pmod4$; $-b$ if $b equiv 3 pmod4$; $2ib$ if $b equiv 0, a equiv 1 pmod4$; and $-2ib$ if $b equiv 0, a equiv 3 pmod4$.
So, if we do the standard major/minor arc decomposition with, say, a bubble around $fracab$ of size $frac1N$, say, then the major arc contribution will be roughly $$fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 1 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))-fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 3 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))$$ plus two analogous terms, for the cases $b equiv 0 pmod4$.
answered 7 hours ago
mathworker21
6,3381727
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 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
0
down vote
Here are some relevant computations. Fix $N$ large. We are interested in $$int_0^1 P_N(alpha)S_N(alpha)^2e(-Nalpha)dalpha$$ where $$P_N(alpha) := sum_p le N e(palpha)$$ $$S_N(alpha) := sum_m le sqrtN e(m^2alpha)$$ $$e(beta) := e^2pi i beta.$$ If $alpha = fracab$ for $b le sqrtlog N$ (say), then $$P_N(alpha) = sum_p le N e(pfracab) = sum_j=0^b-1 sum_substackp le N \ pa equiv j pmodb e(fracjb) approx sum_j=0^b-1 e(fracjb) fracpi(N)phi(b) = fracmu(b)phi(b)pi(N).$$
And, $S_N(fracab) approx fracsqrtNbsum_m le b e(m^2fracab)$. By exact values for Guass sums, the square of the sum is $0$ if $b equiv 2 pmod4$; $b$ if $b equiv 1 pmod4$; $-b$ if $b equiv 3 pmod4$; $2ib$ if $b equiv 0, a equiv 1 pmod4$; and $-2ib$ if $b equiv 0, a equiv 3 pmod4$.
So, if we do the standard major/minor arc decomposition with, say, a bubble around $fracab$ of size $frac1N$, say, then the major arc contribution will be roughly $$fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 1 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))-fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 3 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))$$ plus two analogous terms, for the cases $b equiv 0 pmod4$.
answered 7 hours ago
mathworker21
6,3381727
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
add a comment |Â
up vote
0
down vote
Here are some relevant computations. Fix $N$ large. We are interested in $$int_0^1 P_N(alpha)S_N(alpha)^2e(-Nalpha)dalpha$$ where $$P_N(alpha) := sum_p le N e(palpha)$$ $$S_N(alpha) := sum_m le sqrtN e(m^2alpha)$$ $$e(beta) := e^2pi i beta.$$ If $alpha = fracab$ for $b le sqrtlog N$ (say), then $$P_N(alpha) = sum_p le N e(pfracab) = sum_j=0^b-1 sum_substackp le N \ pa equiv j pmodb e(fracjb) approx sum_j=0^b-1 e(fracjb) fracpi(N)phi(b) = fracmu(b)phi(b)pi(N).$$
And, $S_N(fracab) approx fracsqrtNbsum_m le b e(m^2fracab)$. By exact values for Guass sums, the square of the sum is $0$ if $b equiv 2 pmod4$; $b$ if $b equiv 1 pmod4$; $-b$ if $b equiv 3 pmod4$; $2ib$ if $b equiv 0, a equiv 1 pmod4$; and $-2ib$ if $b equiv 0, a equiv 3 pmod4$.
So, if we do the standard major/minor arc decomposition with, say, a bubble around $fracab$ of size $frac1N$, say, then the major arc contribution will be roughly $$fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 1 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))-fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 3 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))$$ plus two analogous terms, for the cases $b equiv 0 pmod4$.
answered 7 hours ago
mathworker21
6,3381727
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here are some relevant computations. Fix $N$ large. We are interested in $$int_0^1 P_N(alpha)S_N(alpha)^2e(-Nalpha)dalpha$$ where $$P_N(alpha) := sum_p le N e(palpha)$$ $$S_N(alpha) := sum_m le sqrtN e(m^2alpha)$$ $$e(beta) := e^2pi i beta.$$ If $alpha = fracab$ for $b le sqrtlog N$ (say), then $$P_N(alpha) = sum_p le N e(pfracab) = sum_j=0^b-1 sum_substackp le N \ pa equiv j pmodb e(fracjb) approx sum_j=0^b-1 e(fracjb) fracpi(N)phi(b) = fracmu(b)phi(b)pi(N).$$
And, $S_N(fracab) approx fracsqrtNbsum_m le b e(m^2fracab)$. By exact values for Guass sums, the square of the sum is $0$ if $b equiv 2 pmod4$; $b$ if $b equiv 1 pmod4$; $-b$ if $b equiv 3 pmod4$; $2ib$ if $b equiv 0, a equiv 1 pmod4$; and $-2ib$ if $b equiv 0, a equiv 3 pmod4$.
So, if we do the standard major/minor arc decomposition with, say, a bubble around $fracab$ of size $frac1N$, say, then the major arc contribution will be roughly $$fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 1 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))-fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 3 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))$$ plus two analogous terms, for the cases $b equiv 0 pmod4$.
answered 7 hours ago
mathworker21
6,3381727
Here are some relevant computations. Fix $N$ large. We are interested in $$int_0^1 P_N(alpha)S_N(alpha)^2e(-Nalpha)dalpha$$ where $$P_N(alpha) := sum_p le N e(palpha)$$ $$S_N(alpha) := sum_m le sqrtN e(m^2alpha)$$ $$e(beta) := e^2pi i beta.$$ If $alpha = fracab$ for $b le sqrtlog N$ (say), then $$P_N(alpha) = sum_p le N e(pfracab) = sum_j=0^b-1 sum_substackp le N \ pa equiv j pmodb e(fracjb) approx sum_j=0^b-1 e(fracjb) fracpi(N)phi(b) = fracmu(b)phi(b)pi(N).$$
And, $S_N(fracab) approx fracsqrtNbsum_m le b e(m^2fracab)$. By exact values for Guass sums, the square of the sum is $0$ if $b equiv 2 pmod4$; $b$ if $b equiv 1 pmod4$; $-b$ if $b equiv 3 pmod4$; $2ib$ if $b equiv 0, a equiv 1 pmod4$; and $-2ib$ if $b equiv 0, a equiv 3 pmod4$.
So, if we do the standard major/minor arc decomposition with, say, a bubble around $fracab$ of size $frac1N$, say, then the major arc contribution will be roughly $$fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 1 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))-fracNlog Nsum_substack1 le b le sqrtlog N \ b equiv 3 pmod4 fracmu(b)bfracmu(b/(b,N))phi(b/(b,N))$$ plus two analogous terms, for the cases $b equiv 0 pmod4$.
answered 7 hours ago
mathworker21
6,3381727
edited 31 mins ago
answered 7 hours ago
mathworker21
6,3381727
answered 7 hours ago
mathworker21
6,3381727
answered 7 hours ago
mathworker21
6,3381727
6,3381727
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
add a comment |Â
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
Where can I read about this method?
– Lehs
34 mins ago
Where can I read about this method?
– Lehs
34 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
It's called the "circle method". Googling will give more than enough results.
– mathworker21
32 mins ago
add a comment |Â
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1
'two squares of integers' Can you please elaborate? Do you mean $a^2b^2$ or $a^2+b^2$?
– TheSimpliFire
18 hours ago
What do major and minor arcs look like?
– mathworker21
18 hours ago
1
For $n=p+x^2y^2$, $n=37$ does not satisfy this.
– TheSimpliFire
18 hours ago
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
14 hours ago
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
12 hours ago