Prove that every even perfect number is a triangular number.












1












$begingroup$


I know a triangular number is given by the formula $frac{n(n+1)}{2}$



I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.



Please help me to prove this.






elementary-number-theory prime-numbers







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$endgroup$

















    1












    $begingroup$


    I know a triangular number is given by the formula $frac{n(n+1)}{2}$



    I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.



    Please help me to prove this.






    elementary-number-theory prime-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I know a triangular number is given by the formula $frac{n(n+1)}{2}$



      I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.



      Please help me to prove this.






      elementary-number-theory prime-numbers







      share|cite|improve this question











      $endgroup$




      I know a triangular number is given by the formula $frac{n(n+1)}{2}$



      I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.



      Please help me to prove this.






      elementary-number-theory prime-numbers




      elementary-number-theory prime-numbers






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      share|cite|improve this question













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      edited 1 hour ago









      Anirban Niloy

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      666218










      asked 3 hours ago









      Jake GJake G

      442




      442






















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          $begingroup$

          You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.






          share|cite|improve this answer









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            $begingroup$

            You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.






                share|cite|improve this answer









                $endgroup$



                You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Ross MillikanRoss Millikan

                298k23198371




                298k23198371






























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