Sum and number of divisors












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$begingroup$


If the sum of divisors is prime, how can I show that the number of divisors is also prime?
I've tried to use definitons, but I'm not getting anything.






elementary-number-theory







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$endgroup$












  • $begingroup$
    Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 4:58










  • $begingroup$
    Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 5:07










  • $begingroup$
    @WillJagy think Mersenne!
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:10










  • $begingroup$
    Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
    $endgroup$
    – Uday Khanna
    Dec 21 '18 at 6:37










  • $begingroup$
    @LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 18:29


















0












$begingroup$


If the sum of divisors is prime, how can I show that the number of divisors is also prime?
I've tried to use definitons, but I'm not getting anything.






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 4:58










  • $begingroup$
    Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 5:07










  • $begingroup$
    @WillJagy think Mersenne!
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:10










  • $begingroup$
    Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
    $endgroup$
    – Uday Khanna
    Dec 21 '18 at 6:37










  • $begingroup$
    @LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 18:29
















0












0








0





$begingroup$


If the sum of divisors is prime, how can I show that the number of divisors is also prime?
I've tried to use definitons, but I'm not getting anything.






elementary-number-theory







share|cite|improve this question











$endgroup$




If the sum of divisors is prime, how can I show that the number of divisors is also prime?
I've tried to use definitons, but I'm not getting anything.






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 21 '18 at 6:04









Brahadeesh

6,42442363




6,42442363










asked Dec 21 '18 at 4:51









LowkeyLowkey

728




728












  • $begingroup$
    Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 4:58










  • $begingroup$
    Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 5:07










  • $begingroup$
    @WillJagy think Mersenne!
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:10










  • $begingroup$
    Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
    $endgroup$
    – Uday Khanna
    Dec 21 '18 at 6:37










  • $begingroup$
    @LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 18:29




















  • $begingroup$
    Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 4:58










  • $begingroup$
    Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 5:07










  • $begingroup$
    @WillJagy think Mersenne!
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:10










  • $begingroup$
    Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
    $endgroup$
    – Uday Khanna
    Dec 21 '18 at 6:37










  • $begingroup$
    @LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
    $endgroup$
    – Will Jagy
    Dec 21 '18 at 18:29


















$begingroup$
Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 4:58




$begingroup$
Do you mean that "if $sigma(n)$, the sum of all divisors of $n$, is prime, then $d(n)$, the number of divisors of $n$, is prime"?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 4:58












$begingroup$
Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
$endgroup$
– Will Jagy
Dec 21 '18 at 5:07




$begingroup$
Do you have any examples of numbers $n$ such that $sigma(n)$ is prime? I can think of only one such $n.$
$endgroup$
– Will Jagy
Dec 21 '18 at 5:07












$begingroup$
@WillJagy think Mersenne!
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:10




$begingroup$
@WillJagy think Mersenne!
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:10












$begingroup$
Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
$endgroup$
– Uday Khanna
Dec 21 '18 at 6:37




$begingroup$
Clearly the numbers with an odd no. of divisors are perfect squares. So we have to show if the sum of the divisors of a perfect square are prime then the no. of divisors is also prime.
$endgroup$
– Uday Khanna
Dec 21 '18 at 6:37












$begingroup$
@LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
$endgroup$
– Will Jagy
Dec 21 '18 at 18:29






$begingroup$
@LordSharktheUnknown good, found first few 2 = 2 sigma 3 d 2 /// 4 = 2^2 sigma 7 d 3 /// 9 = 3^2 sigma 13 d 3 /// 16 = 2^4 sigma 31 d 5 /// 25 = 5^2 sigma 31 d 3 /// 64 = 2^6 sigma 127 d 7 /// 289 = 17^2 sigma 307 d 3 /// 729 = 3^6 sigma 1093 d 7
$endgroup$
– Will Jagy
Dec 21 '18 at 18:29












1 Answer
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The sum of the divisiors of the number $$N=p_1^{a_1}cdots p_n^{a_n}$$ is $$(1+p_1+cdots +p_1^{a_1})cdots (1+p_n+cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1mid p^{n+1}-1$ giving the non-trivial factor $frac{p^a-1}{p-1}$. This completes the proof of the statement.






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    $begingroup$

    The sum of the divisiors of the number $$N=p_1^{a_1}cdots p_n^{a_n}$$ is $$(1+p_1+cdots +p_1^{a_1})cdots (1+p_n+cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1mid p^{n+1}-1$ giving the non-trivial factor $frac{p^a-1}{p-1}$. This completes the proof of the statement.






    share|cite|improve this answer









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      1












      $begingroup$

      The sum of the divisiors of the number $$N=p_1^{a_1}cdots p_n^{a_n}$$ is $$(1+p_1+cdots +p_1^{a_1})cdots (1+p_n+cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1mid p^{n+1}-1$ giving the non-trivial factor $frac{p^a-1}{p-1}$. This completes the proof of the statement.






      share|cite|improve this answer









      $endgroup$
















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        $begingroup$

        The sum of the divisiors of the number $$N=p_1^{a_1}cdots p_n^{a_n}$$ is $$(1+p_1+cdots +p_1^{a_1})cdots (1+p_n+cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1mid p^{n+1}-1$ giving the non-trivial factor $frac{p^a-1}{p-1}$. This completes the proof of the statement.






        share|cite|improve this answer









        $endgroup$



        The sum of the divisiors of the number $$N=p_1^{a_1}cdots p_n^{a_n}$$ is $$(1+p_1+cdots +p_1^{a_1})cdots (1+p_n+cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1mid p^{n+1}-1$ giving the non-trivial factor $frac{p^a-1}{p-1}$. This completes the proof of the statement.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 7:22









        PeterPeter

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