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











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    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


















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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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