Prove that every even perfect number is a triangular number.












4












$begingroup$


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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.



Please help me to prove this.






elementary-number-theory prime-numbers







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.



    Please help me to prove this.






    elementary-number-theory prime-numbers







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      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 $2^{n-1}(2^{n}-1)$ 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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.



      Please help me to prove this.






      elementary-number-theory prime-numbers




      elementary-number-theory prime-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 4 at 8:40









      Vinyl_cape_jawa

      3,33011433




      3,33011433










      asked Mar 4 at 3:58









      Jake GJake G

      656




      656






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          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. Therefore $n=2^{k}-1$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            n should be 2^(k)-1
            $endgroup$
            – Jake G
            Mar 4 at 8:00










          • $begingroup$
            @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
            $endgroup$
            – Vinyl_cape_jawa
            Mar 4 at 9:30










          • $begingroup$
            @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
            $endgroup$
            – David Mulder
            Mar 4 at 10:33





















          3












          $begingroup$

          You could think as follows:



          Every triangular number you can represent as follows:



          $$
          bullet\
          bullet bullet\
          bullet bullet bullet \
          ...
          $$



          that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.



          Multiplying this by $2$ gives a rectangle



          $$
          bullet circ circ circ\
          bullet bullet circ circ\
          bullet bullet bullet circ\
          $$



          with one side exactly one unit longer than the other.



          So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.



          Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed



          $$
          frac{n(n+1)}{2}=2^{k-1}(2^k-1),
          $$



          so
          $$
          text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
          $$






          share|cite|improve this answer











          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            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. Therefore $n=2^{k}-1$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              n should be 2^(k)-1
              $endgroup$
              – Jake G
              Mar 4 at 8:00










            • $begingroup$
              @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
              $endgroup$
              – Vinyl_cape_jawa
              Mar 4 at 9:30










            • $begingroup$
              @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
              $endgroup$
              – David Mulder
              Mar 4 at 10:33


















            6












            $begingroup$

            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. Therefore $n=2^{k}-1$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              n should be 2^(k)-1
              $endgroup$
              – Jake G
              Mar 4 at 8:00










            • $begingroup$
              @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
              $endgroup$
              – Vinyl_cape_jawa
              Mar 4 at 9:30










            • $begingroup$
              @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
              $endgroup$
              – David Mulder
              Mar 4 at 10:33
















            6












            6








            6





            $begingroup$

            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. Therefore $n=2^{k}-1$






            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. Therefore $n=2^{k}-1$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 4 at 8:48









            Jake G

            656




            656










            answered Mar 4 at 4:04









            Ross MillikanRoss Millikan

            301k24200375




            301k24200375












            • $begingroup$
              n should be 2^(k)-1
              $endgroup$
              – Jake G
              Mar 4 at 8:00










            • $begingroup$
              @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
              $endgroup$
              – Vinyl_cape_jawa
              Mar 4 at 9:30










            • $begingroup$
              @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
              $endgroup$
              – David Mulder
              Mar 4 at 10:33




















            • $begingroup$
              n should be 2^(k)-1
              $endgroup$
              – Jake G
              Mar 4 at 8:00










            • $begingroup$
              @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
              $endgroup$
              – Vinyl_cape_jawa
              Mar 4 at 9:30










            • $begingroup$
              @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
              $endgroup$
              – David Mulder
              Mar 4 at 10:33


















            $begingroup$
            n should be 2^(k)-1
            $endgroup$
            – Jake G
            Mar 4 at 8:00




            $begingroup$
            n should be 2^(k)-1
            $endgroup$
            – Jake G
            Mar 4 at 8:00












            $begingroup$
            @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
            $endgroup$
            – Vinyl_cape_jawa
            Mar 4 at 9:30




            $begingroup$
            @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
            $endgroup$
            – Vinyl_cape_jawa
            Mar 4 at 9:30












            $begingroup$
            @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
            $endgroup$
            – David Mulder
            Mar 4 at 10:33






            $begingroup$
            @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
            $endgroup$
            – David Mulder
            Mar 4 at 10:33













            3












            $begingroup$

            You could think as follows:



            Every triangular number you can represent as follows:



            $$
            bullet\
            bullet bullet\
            bullet bullet bullet \
            ...
            $$



            that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.



            Multiplying this by $2$ gives a rectangle



            $$
            bullet circ circ circ\
            bullet bullet circ circ\
            bullet bullet bullet circ\
            $$



            with one side exactly one unit longer than the other.



            So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.



            Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed



            $$
            frac{n(n+1)}{2}=2^{k-1}(2^k-1),
            $$



            so
            $$
            text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
            $$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              You could think as follows:



              Every triangular number you can represent as follows:



              $$
              bullet\
              bullet bullet\
              bullet bullet bullet \
              ...
              $$



              that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.



              Multiplying this by $2$ gives a rectangle



              $$
              bullet circ circ circ\
              bullet bullet circ circ\
              bullet bullet bullet circ\
              $$



              with one side exactly one unit longer than the other.



              So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.



              Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed



              $$
              frac{n(n+1)}{2}=2^{k-1}(2^k-1),
              $$



              so
              $$
              text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
              $$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                You could think as follows:



                Every triangular number you can represent as follows:



                $$
                bullet\
                bullet bullet\
                bullet bullet bullet \
                ...
                $$



                that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.



                Multiplying this by $2$ gives a rectangle



                $$
                bullet circ circ circ\
                bullet bullet circ circ\
                bullet bullet bullet circ\
                $$



                with one side exactly one unit longer than the other.



                So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.



                Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed



                $$
                frac{n(n+1)}{2}=2^{k-1}(2^k-1),
                $$



                so
                $$
                text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
                $$






                share|cite|improve this answer











                $endgroup$



                You could think as follows:



                Every triangular number you can represent as follows:



                $$
                bullet\
                bullet bullet\
                bullet bullet bullet \
                ...
                $$



                that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.



                Multiplying this by $2$ gives a rectangle



                $$
                bullet circ circ circ\
                bullet bullet circ circ\
                bullet bullet bullet circ\
                $$



                with one side exactly one unit longer than the other.



                So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.



                Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed



                $$
                frac{n(n+1)}{2}=2^{k-1}(2^k-1),
                $$



                so
                $$
                text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 4 at 9:37

























                answered Mar 4 at 9:20









                Vinyl_cape_jawaVinyl_cape_jawa

                3,33011433




                3,33011433






























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