Sum of divisors of perfect square












2












$begingroup$


Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you seen this?
    $endgroup$
    – Don Thousand
    Dec 24 '18 at 19:19










  • $begingroup$
    Thank you,I don't know if that helps me
    $endgroup$
    – argamon
    Dec 24 '18 at 19:21










  • $begingroup$
    @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:08












  • $begingroup$
    I bet you had tried that, not that interesting..
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:12










  • $begingroup$
    Haha, it's just more useful if I know in general but I greatly appreciate your help
    $endgroup$
    – argamon
    Dec 24 '18 at 20:13
















2












$begingroup$


Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you seen this?
    $endgroup$
    – Don Thousand
    Dec 24 '18 at 19:19










  • $begingroup$
    Thank you,I don't know if that helps me
    $endgroup$
    – argamon
    Dec 24 '18 at 19:21










  • $begingroup$
    @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:08












  • $begingroup$
    I bet you had tried that, not that interesting..
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:12










  • $begingroup$
    Haha, it's just more useful if I know in general but I greatly appreciate your help
    $endgroup$
    – argamon
    Dec 24 '18 at 20:13














2












2








2





$begingroup$


Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$




Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 19:46







argamon

















asked Dec 24 '18 at 19:08









argamonargamon

756




756












  • $begingroup$
    Have you seen this?
    $endgroup$
    – Don Thousand
    Dec 24 '18 at 19:19










  • $begingroup$
    Thank you,I don't know if that helps me
    $endgroup$
    – argamon
    Dec 24 '18 at 19:21










  • $begingroup$
    @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:08












  • $begingroup$
    I bet you had tried that, not that interesting..
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:12










  • $begingroup$
    Haha, it's just more useful if I know in general but I greatly appreciate your help
    $endgroup$
    – argamon
    Dec 24 '18 at 20:13


















  • $begingroup$
    Have you seen this?
    $endgroup$
    – Don Thousand
    Dec 24 '18 at 19:19










  • $begingroup$
    Thank you,I don't know if that helps me
    $endgroup$
    – argamon
    Dec 24 '18 at 19:21










  • $begingroup$
    @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:08












  • $begingroup$
    I bet you had tried that, not that interesting..
    $endgroup$
    – rsadhvika
    Dec 24 '18 at 20:12










  • $begingroup$
    Haha, it's just more useful if I know in general but I greatly appreciate your help
    $endgroup$
    – argamon
    Dec 24 '18 at 20:13
















$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19




$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19












$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21




$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21












$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08






$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08














$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12




$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12












$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13




$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13










2 Answers
2






active

oldest

votes


















4












$begingroup$

It looks to me like $sigma(627^2)=sigma(749^2)$.



Feel free to check my work. (Or Python's work!)



And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$



    and so on.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
      $endgroup$
      – argamon
      Dec 24 '18 at 19:45










    • $begingroup$
      I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
      $endgroup$
      – lulu
      Dec 24 '18 at 19:48












    • $begingroup$
      In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
      $endgroup$
      – Ross Millikan
      Dec 24 '18 at 20:22










    • $begingroup$
      You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
      $endgroup$
      – Ross Millikan
      Dec 24 '18 at 20:44












    • $begingroup$
      @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
      $endgroup$
      – lulu
      Dec 24 '18 at 20:52











    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%2f3051553%2fsum-of-divisors-of-perfect-square%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    It looks to me like $sigma(627^2)=sigma(749^2)$.



    Feel free to check my work. (Or Python's work!)



    And here's an even-even pair:
    $sigma(740^2)=sigma(878^2)$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      It looks to me like $sigma(627^2)=sigma(749^2)$.



      Feel free to check my work. (Or Python's work!)



      And here's an even-even pair:
      $sigma(740^2)=sigma(878^2)$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        It looks to me like $sigma(627^2)=sigma(749^2)$.



        Feel free to check my work. (Or Python's work!)



        And here's an even-even pair:
        $sigma(740^2)=sigma(878^2)$.






        share|cite|improve this answer











        $endgroup$



        It looks to me like $sigma(627^2)=sigma(749^2)$.



        Feel free to check my work. (Or Python's work!)



        And here's an even-even pair:
        $sigma(740^2)=sigma(878^2)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 20:41

























        answered Dec 24 '18 at 20:33









        paw88789paw88789

        29.1k12349




        29.1k12349























            4












            $begingroup$

            As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$



            and so on.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
              $endgroup$
              – argamon
              Dec 24 '18 at 19:45










            • $begingroup$
              I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
              $endgroup$
              – lulu
              Dec 24 '18 at 19:48












            • $begingroup$
              In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:22










            • $begingroup$
              You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:44












            • $begingroup$
              @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
              $endgroup$
              – lulu
              Dec 24 '18 at 20:52
















            4












            $begingroup$

            As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$



            and so on.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
              $endgroup$
              – argamon
              Dec 24 '18 at 19:45










            • $begingroup$
              I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
              $endgroup$
              – lulu
              Dec 24 '18 at 19:48












            • $begingroup$
              In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:22










            • $begingroup$
              You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:44












            • $begingroup$
              @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
              $endgroup$
              – lulu
              Dec 24 '18 at 20:52














            4












            4








            4





            $begingroup$

            As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$



            and so on.






            share|cite|improve this answer











            $endgroup$



            As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$



            and so on.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 24 '18 at 20:54


























            community wiki





            2 revs
            lulu













            • $begingroup$
              That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
              $endgroup$
              – argamon
              Dec 24 '18 at 19:45










            • $begingroup$
              I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
              $endgroup$
              – lulu
              Dec 24 '18 at 19:48












            • $begingroup$
              In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:22










            • $begingroup$
              You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:44












            • $begingroup$
              @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
              $endgroup$
              – lulu
              Dec 24 '18 at 20:52


















            • $begingroup$
              That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
              $endgroup$
              – argamon
              Dec 24 '18 at 19:45










            • $begingroup$
              I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
              $endgroup$
              – lulu
              Dec 24 '18 at 19:48












            • $begingroup$
              In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:22










            • $begingroup$
              You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
              $endgroup$
              – Ross Millikan
              Dec 24 '18 at 20:44












            • $begingroup$
              @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
              $endgroup$
              – lulu
              Dec 24 '18 at 20:52
















            $begingroup$
            That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
            $endgroup$
            – argamon
            Dec 24 '18 at 19:45




            $begingroup$
            That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
            $endgroup$
            – argamon
            Dec 24 '18 at 19:45












            $begingroup$
            I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
            $endgroup$
            – lulu
            Dec 24 '18 at 19:48






            $begingroup$
            I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
            $endgroup$
            – lulu
            Dec 24 '18 at 19:48














            $begingroup$
            In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
            $endgroup$
            – Ross Millikan
            Dec 24 '18 at 20:22




            $begingroup$
            In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
            $endgroup$
            – Ross Millikan
            Dec 24 '18 at 20:22












            $begingroup$
            You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
            $endgroup$
            – Ross Millikan
            Dec 24 '18 at 20:44






            $begingroup$
            You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
            $endgroup$
            – Ross Millikan
            Dec 24 '18 at 20:44














            $begingroup$
            @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
            $endgroup$
            – lulu
            Dec 24 '18 at 20:52




            $begingroup$
            @RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
            $endgroup$
            – lulu
            Dec 24 '18 at 20:52


















            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%2f3051553%2fsum-of-divisors-of-perfect-square%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

            If I really need a card on my start hand, how many mulligans make sense? [duplicate]

            Image
            5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
            Read more

            Alcedinidae

            Image
            Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
            Read more

            Can an atomic nucleus contain both particles and antiparticles? [duplicate]

            Image
            9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
            Read more