What are some equivalent statements of (strong) Goldbach Conjecture?
$begingroup$
What are some equivalent statements of (strong) Goldbach Conjecture ?
We all know that Riemann Hypothesis has some interesting equivalent statements.
My favorites are involved with Mertens function, error terms of Prime Number Theorem, and Farey sequences. Those equivalent statements do not use Riemann Zeta function directly, but provide additional insights about Riemann Hypothesis from very different angles.
What are some equivalent statements of (strong) Goldbach Conjecture ?
to shed lights on this problems from different angles ?
number-theory algebraic-number-theory analytic-number-theory riemann-hypothesis goldbachs-conjecture
$endgroup$
add a comment |
$begingroup$
What are some equivalent statements of (strong) Goldbach Conjecture ?
We all know that Riemann Hypothesis has some interesting equivalent statements.
My favorites are involved with Mertens function, error terms of Prime Number Theorem, and Farey sequences. Those equivalent statements do not use Riemann Zeta function directly, but provide additional insights about Riemann Hypothesis from very different angles.
What are some equivalent statements of (strong) Goldbach Conjecture ?
to shed lights on this problems from different angles ?
number-theory algebraic-number-theory analytic-number-theory riemann-hypothesis goldbachs-conjecture
$endgroup$
add a comment |
$begingroup$
What are some equivalent statements of (strong) Goldbach Conjecture ?
We all know that Riemann Hypothesis has some interesting equivalent statements.
My favorites are involved with Mertens function, error terms of Prime Number Theorem, and Farey sequences. Those equivalent statements do not use Riemann Zeta function directly, but provide additional insights about Riemann Hypothesis from very different angles.
What are some equivalent statements of (strong) Goldbach Conjecture ?
to shed lights on this problems from different angles ?
number-theory algebraic-number-theory analytic-number-theory riemann-hypothesis goldbachs-conjecture
$endgroup$
What are some equivalent statements of (strong) Goldbach Conjecture ?
We all know that Riemann Hypothesis has some interesting equivalent statements.
My favorites are involved with Mertens function, error terms of Prime Number Theorem, and Farey sequences. Those equivalent statements do not use Riemann Zeta function directly, but provide additional insights about Riemann Hypothesis from very different angles.
What are some equivalent statements of (strong) Goldbach Conjecture ?
to shed lights on this problems from different angles ?
number-theory algebraic-number-theory analytic-number-theory riemann-hypothesis goldbachs-conjecture
number-theory algebraic-number-theory analytic-number-theory riemann-hypothesis goldbachs-conjecture
asked Dec 29 '14 at 16:51
daviddavid
1092
1092
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
For every integer $n geq 2$ there exist integers $k, p$ and $q$ with $0 leq k leq n-2$
and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)
$endgroup$
add a comment |
$begingroup$
For every integer $n geq 1$ there exists primes $p$ and $q$ such that $varphi(p)+varphi(q)=2n$
where $varphi$ is Euler's Totient function .
$endgroup$
add a comment |
$begingroup$
"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"
... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"
$endgroup$
add a comment |
$begingroup$
Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), nge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.
It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.
It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.
$endgroup$
add a comment |
$begingroup$
Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3leq p_i<m<p_j<2m$, let $n=prod{p_j}$. Let $c_i=2m-p_i$, and $a=prod{c_i}$. Then $left(frac{a}{n}right)=0$ where $left(frac{a}{n}right)$ is the Jacobi symbol.
$endgroup$
add a comment |
$begingroup$
I conjectured this:
Every integer $n>3$ is halfway between $2$ primes.
See the proof
here
.
By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!
$endgroup$
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1084743%2fwhat-are-some-equivalent-statements-of-strong-goldbach-conjecture%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For every integer $n geq 2$ there exist integers $k, p$ and $q$ with $0 leq k leq n-2$
and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)
$endgroup$
add a comment |
$begingroup$
For every integer $n geq 2$ there exist integers $k, p$ and $q$ with $0 leq k leq n-2$
and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)
$endgroup$
add a comment |
$begingroup$
For every integer $n geq 2$ there exist integers $k, p$ and $q$ with $0 leq k leq n-2$
and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)
$endgroup$
For every integer $n geq 2$ there exist integers $k, p$ and $q$ with $0 leq k leq n-2$
and with $p$ and $q$ prime such that $n^2 - k^2 = pq$. (http://www.maa.org/sites/default/files/Reformulation-Gerstein20557.pdf)
answered Jan 16 '15 at 22:06
Loreno HeerLoreno Heer
3,34411534
3,34411534
add a comment |
add a comment |
$begingroup$
For every integer $n geq 1$ there exists primes $p$ and $q$ such that $varphi(p)+varphi(q)=2n$
where $varphi$ is Euler's Totient function .
$endgroup$
add a comment |
$begingroup$
For every integer $n geq 1$ there exists primes $p$ and $q$ such that $varphi(p)+varphi(q)=2n$
where $varphi$ is Euler's Totient function .
$endgroup$
add a comment |
$begingroup$
For every integer $n geq 1$ there exists primes $p$ and $q$ such that $varphi(p)+varphi(q)=2n$
where $varphi$ is Euler's Totient function .
$endgroup$
For every integer $n geq 1$ there exists primes $p$ and $q$ such that $varphi(p)+varphi(q)=2n$
where $varphi$ is Euler's Totient function .
answered Dec 29 '14 at 18:03
Peđa TerzićPeđa Terzić
7,86522572
7,86522572
add a comment |
add a comment |
$begingroup$
"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"
... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"
$endgroup$
add a comment |
$begingroup$
"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"
... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"
$endgroup$
add a comment |
$begingroup$
"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"
... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"
$endgroup$
"Every natural number greater than one can be written as the arithmetical mean of two primes (not necessarily different)"
... from my point of view, it simplifies the restrictions (to be "greater than one" is simpler than to be "even and greater than two"); and it is intuitively better, as "arithmetic mean" gives a constructive recipe to obtain the number as the center of their associated "prime couples"
edited Mar 6 '18 at 11:16
answered Nov 19 '17 at 13:12
Perspectiva8Perspectiva8
16212
16212
add a comment |
add a comment |
$begingroup$
Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), nge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.
It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.
It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.
$endgroup$
add a comment |
$begingroup$
Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), nge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.
It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.
It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.
$endgroup$
add a comment |
$begingroup$
Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), nge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.
It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.
It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.
$endgroup$
Reformulating the answer of @LorenoHeer, using the fact that the sum of the first $n$ odd numbers is equal to $n^2$: Every term $(2n-1), nge4$ of the sequence of odd numbers is the final term of at least one series containing a prime number $p$ consecutive terms, the middle term of which is a prime number $q$.
It can be seen that the sum of such a series is $pq$, that is, the number of terms $(p)$ times the average value of the terms $(q)$. The series is also the sum of the first $n$ odd numbers $(=n^2)$ less the sum of the first $k$ odd numbers $(=k^2)$, or $n^2-k^2$, as given in the referenced answer. Since $(2n-1)$ is the $n^{th}$ odd number and there are $p$ terms in the series, $k=n-p$.
It is unsurprising that one can pick an odd prime and make it the center term in a series of consecutive odd numbers with a prime number of terms. What strikes me is that if one does so systematically, the set of final terms of all such series saturates the (suitably large) odd numbers. Tangentially, this says something about the distribution of prime numbers; whether what it says is distinct from what the canonical formulation of the Goldbach conjecture says about the distribution of prime numbers is beyond my ken.
edited Jul 12 '18 at 15:14
answered Jul 12 '18 at 2:27
Keith BackmanKeith Backman
1,2691812
1,2691812
add a comment |
add a comment |
$begingroup$
Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3leq p_i<m<p_j<2m$, let $n=prod{p_j}$. Let $c_i=2m-p_i$, and $a=prod{c_i}$. Then $left(frac{a}{n}right)=0$ where $left(frac{a}{n}right)$ is the Jacobi symbol.
$endgroup$
add a comment |
$begingroup$
Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3leq p_i<m<p_j<2m$, let $n=prod{p_j}$. Let $c_i=2m-p_i$, and $a=prod{c_i}$. Then $left(frac{a}{n}right)=0$ where $left(frac{a}{n}right)$ is the Jacobi symbol.
$endgroup$
add a comment |
$begingroup$
Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3leq p_i<m<p_j<2m$, let $n=prod{p_j}$. Let $c_i=2m-p_i$, and $a=prod{c_i}$. Then $left(frac{a}{n}right)=0$ where $left(frac{a}{n}right)$ is the Jacobi symbol.
$endgroup$
Here is a more abstruse equivalent to the Goldbach conjecture. Consider (sufficiently large) even numbers $2m$. If $m$ is prime, the conjecture is true, so we omit that possibility from further consideration. For odd prime numbers $3leq p_i<m<p_j<2m$, let $n=prod{p_j}$. Let $c_i=2m-p_i$, and $a=prod{c_i}$. Then $left(frac{a}{n}right)=0$ where $left(frac{a}{n}right)$ is the Jacobi symbol.
edited Jul 20 '18 at 17:37
answered Jul 20 '18 at 17:22
Keith BackmanKeith Backman
1,2691812
1,2691812
add a comment |
add a comment |
$begingroup$
I conjectured this:
Every integer $n>3$ is halfway between $2$ primes.
See the proof
here
.
By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!
$endgroup$
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
add a comment |
$begingroup$
I conjectured this:
Every integer $n>3$ is halfway between $2$ primes.
See the proof
here
.
By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!
$endgroup$
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
add a comment |
$begingroup$
I conjectured this:
Every integer $n>3$ is halfway between $2$ primes.
See the proof
here
.
By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!
$endgroup$
I conjectured this:
Every integer $n>3$ is halfway between $2$ primes.
See the proof
here
.
By the way, this is a very good question. Like you said, the best way to solve this old problem must certainly be to look at it from new angles. There should be a lot more answers here. I have 2-3 other statements in mind. I'll do some searching and post them soon!
answered Jan 5 at 1:57
François HuppéFrançois Huppé
362111
362111
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
add a comment |
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
1
1
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
That the RH is equivalent to $pi(x)-Li(x)= O(x^{1/2+epsilon})$ is far for being trivial (the proof is quite the same as the PNT). Whereas "Every integer n>3 is halfway between 2 primes" is trivially the Goldbach conjecture.
$endgroup$
– reuns
Jan 5 at 2:44
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
$begingroup$
For $2$ conjectures A and B, if A is trivially B, can we say A is equivalent to B ? My understanding of "New angles" may not be accurate. Also, It seems like the case of an even integer $2n$ where $n$ is prime respects Goldbach conjecture but not mine...
$endgroup$
– François Huppé
Jan 5 at 3:30
1
1
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
$begingroup$
$p$ is halfway between $p$ and $p$. That $2n = p+q$ is the same as $n = frac{p+q}{2}$ which is the same as $p = n+a, q = n-a$ (with $ a = p-n$).
$endgroup$
– reuns
Jan 5 at 4:03
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1084743%2fwhat-are-some-equivalent-statements-of-strong-goldbach-conjecture%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown