What are some equivalent statements of (strong) Goldbach Conjecture?












12












$begingroup$


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 ?










share|cite|improve this question









$endgroup$

















    12












    $begingroup$


    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 ?










    share|cite|improve this question









    $endgroup$















      12












      12








      12


      4



      $begingroup$


      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 ?










      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 29 '14 at 16:51









      daviddavid

      1092




      1092






















          6 Answers
          6






          active

          oldest

          votes


















          3












          $begingroup$

          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)






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            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 .






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              "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"






              share|cite|improve this answer











              $endgroup$





















                0












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$





















                  0












                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$





















                    0












                    $begingroup$

                    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!






                    share|cite|improve this answer









                    $endgroup$









                    • 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













                    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
                    });


                    }
                    });














                    draft saved

                    draft discarded


















                    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









                    3












                    $begingroup$

                    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)






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      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)






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        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)






                        share|cite|improve this answer









                        $endgroup$



                        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)







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 16 '15 at 22:06









                        Loreno HeerLoreno Heer

                        3,34411534




                        3,34411534























                            2












                            $begingroup$

                            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 .






                            share|cite|improve this answer









                            $endgroup$


















                              2












                              $begingroup$

                              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 .






                              share|cite|improve this answer









                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                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 .






                                share|cite|improve this answer









                                $endgroup$



                                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 .







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 29 '14 at 18:03









                                Peđa TerzićPeđa Terzić

                                7,86522572




                                7,86522572























                                    0












                                    $begingroup$

                                    "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"






                                    share|cite|improve this answer











                                    $endgroup$


















                                      0












                                      $begingroup$

                                      "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"






                                      share|cite|improve this answer











                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        "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"






                                        share|cite|improve this answer











                                        $endgroup$



                                        "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"







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Mar 6 '18 at 11:16

























                                        answered Nov 19 '17 at 13:12









                                        Perspectiva8Perspectiva8

                                        16212




                                        16212























                                            0












                                            $begingroup$

                                            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.






                                            share|cite|improve this answer











                                            $endgroup$


















                                              0












                                              $begingroup$

                                              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.






                                              share|cite|improve this answer











                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                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.






                                                share|cite|improve this answer











                                                $endgroup$



                                                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.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Jul 12 '18 at 15:14

























                                                answered Jul 12 '18 at 2:27









                                                Keith BackmanKeith Backman

                                                1,2691812




                                                1,2691812























                                                    0












                                                    $begingroup$

                                                    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.






                                                    share|cite|improve this answer











                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      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.






                                                      share|cite|improve this answer











                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        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.






                                                        share|cite|improve this answer











                                                        $endgroup$



                                                        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.







                                                        share|cite|improve this answer














                                                        share|cite|improve this answer



                                                        share|cite|improve this answer








                                                        edited Jul 20 '18 at 17:37

























                                                        answered Jul 20 '18 at 17:22









                                                        Keith BackmanKeith Backman

                                                        1,2691812




                                                        1,2691812























                                                            0












                                                            $begingroup$

                                                            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!






                                                            share|cite|improve this answer









                                                            $endgroup$









                                                            • 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


















                                                            0












                                                            $begingroup$

                                                            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!






                                                            share|cite|improve this answer









                                                            $endgroup$









                                                            • 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
















                                                            0












                                                            0








                                                            0





                                                            $begingroup$

                                                            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!






                                                            share|cite|improve this answer









                                                            $endgroup$



                                                            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!







                                                            share|cite|improve this answer












                                                            share|cite|improve this answer



                                                            share|cite|improve this answer










                                                            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
















                                                            • 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




















                                                            draft saved

                                                            draft discarded




















































                                                            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.




                                                            draft saved


                                                            draft discarded














                                                            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





















































                                                            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







                                                            Popular posts from this blog

                                                            Human spaceflight

                                                            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                                            File:DeusFollowingSea.jpg