On a theorem of Zumkeller related to twin primes

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Let $p_1,p_2$ be a twin prime pair, $phi(n)$ denote Euler's totient function and $sigma(n)$ the sum-of-divisors function. Reinhard Zumkeller proved in 2002 that



$$
phi(p_2) = sigma(p_1).
$$



However I cannot find Zumkeller's paper proving this fact. Can anyone kindly point me to a link where I can read his Theorem?










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    Let $p_1,p_2$ be a twin prime pair, $phi(n)$ denote Euler's totient function and $sigma(n)$ the sum-of-divisors function. Reinhard Zumkeller proved in 2002 that



    $$
    phi(p_2) = sigma(p_1).
    $$



    However I cannot find Zumkeller's paper proving this fact. Can anyone kindly point me to a link where I can read his Theorem?










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






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      Let $p_1,p_2$ be a twin prime pair, $phi(n)$ denote Euler's totient function and $sigma(n)$ the sum-of-divisors function. Reinhard Zumkeller proved in 2002 that



      $$
      phi(p_2) = sigma(p_1).
      $$



      However I cannot find Zumkeller's paper proving this fact. Can anyone kindly point me to a link where I can read his Theorem?










      share|cite|improve this question















      Let $p_1,p_2$ be a twin prime pair, $phi(n)$ denote Euler's totient function and $sigma(n)$ the sum-of-divisors function. Reinhard Zumkeller proved in 2002 that



      $$
      phi(p_2) = sigma(p_1).
      $$



      However I cannot find Zumkeller's paper proving this fact. Can anyone kindly point me to a link where I can read his Theorem?







      reference-request prime-numbers divisor-sum prime-twins






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      edited Sep 9 at 11:44









      joriki

      169k10181337




      169k10181337










      asked Sep 9 at 10:30









      PierreTheFermented

      1,3421927




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          There's not much to prove, certainly not enough for a paper. $phi(p_2)=p_2-1$, as for any prime, and $sigma(p_1)=p_1+1$, as for any prime; since they're twin primes, $p_2-1=p_1+1$.






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          • 1




            Oh, indeed :) Thanks!
            – PierreTheFermented
            Sep 9 at 12:33










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          up vote
          4
          down vote



          accepted










          There's not much to prove, certainly not enough for a paper. $phi(p_2)=p_2-1$, as for any prime, and $sigma(p_1)=p_1+1$, as for any prime; since they're twin primes, $p_2-1=p_1+1$.






          share|cite|improve this answer
















          • 1




            Oh, indeed :) Thanks!
            – PierreTheFermented
            Sep 9 at 12:33














          up vote
          4
          down vote



          accepted










          There's not much to prove, certainly not enough for a paper. $phi(p_2)=p_2-1$, as for any prime, and $sigma(p_1)=p_1+1$, as for any prime; since they're twin primes, $p_2-1=p_1+1$.






          share|cite|improve this answer
















          • 1




            Oh, indeed :) Thanks!
            – PierreTheFermented
            Sep 9 at 12:33












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          There's not much to prove, certainly not enough for a paper. $phi(p_2)=p_2-1$, as for any prime, and $sigma(p_1)=p_1+1$, as for any prime; since they're twin primes, $p_2-1=p_1+1$.






          share|cite|improve this answer












          There's not much to prove, certainly not enough for a paper. $phi(p_2)=p_2-1$, as for any prime, and $sigma(p_1)=p_1+1$, as for any prime; since they're twin primes, $p_2-1=p_1+1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 9 at 11:45









          joriki

          169k10181337




          169k10181337







          • 1




            Oh, indeed :) Thanks!
            – PierreTheFermented
            Sep 9 at 12:33












          • 1




            Oh, indeed :) Thanks!
            – PierreTheFermented
            Sep 9 at 12:33







          1




          1




          Oh, indeed :) Thanks!
          – PierreTheFermented
          Sep 9 at 12:33




          Oh, indeed :) Thanks!
          – PierreTheFermented
          Sep 9 at 12:33

















           

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