# Is $n=6$ the only integer satisfies ${\sigma}_x(n) \equiv 0\bmod{n}$ for every odd integer $x > 0$ and $2 (\bmod n)$ if $x$ is even integer? [closed]

After a few computations in wolfram alpha about the divisor function for some values of $n$ to look the behavior of $\sigma_x(n)\bmod n$ for $\,n=6,\,$ i got this result : $\sigma_x(6)=0 \bmod 6$ for $x$ odd and 2 mod 6 if $x$ is even

Edit:01 :${\sigma}_x(n) =\sum_{d|n} d^x$ is the sum divisor function

Note:01:I edited the question just to define the sum divisor function

My question here :

Is $n=6$ the only integer satisfies $\sigma_x(n) \equiv 0\bmod n$ for every odd integer $x > 0$ and $2 \bmod n$ if $x$ is even integer ? and if it is how do i show this ?

Note :02:I want to know more about periodicity of the divisor function

Thank you for any help

• You need to fix these $\LaTeX$ errors. – Włodzimierz Holsztyński Jan 2 '16 at 4:48
• Ok, I fixed $\LaTeX$. Now you need to fix typos, and if you can help it, also English or logic ("logic" in the everyday sense). *** Short sentences would help a lot (as well as avoiding pronouns). – Włodzimierz Holsztyński Jan 2 '16 at 5:33
• You did not compute enough terms. sigma 23 of 6 is twice a square, so sigma 24 of 6 is odd. Gerhard "Check The OEIS Each Time" Paseman, 2016.01.02 – Gerhard Paseman Jan 2 '16 at 9:46
• Also, you need to sharpen your formulation to exclude trivial answers such as: " If x is such a number, so is sigma(sigma(x)).", as well as what evidence you use precisely to base your conjecture (which seems to be not many iterates). It is unclear that sigma possesses any periodicity properties modulo any large integer, and you should perform more tests before posting a question like that in the title. Gerhard "Think Some More Before Asking" Paseman, 2016.01.02. – Gerhard Paseman Jan 2 '16 at 10:02
• @GerhardPaseman, I don't meant iterating sum divisor function ,pleas check this :sigma_23(6)mod6=0 is not twice :wolframalpha.com/input/?i=sigma_23%286%29mod6 – zeraoulia rafik Jan 2 '16 at 20:42

Let $r=\gcd(k,e+1)$, and $p$ a prime. Then $\sigma_k(p^e) \equiv r\frac{p^{e+1}-1}{p^r -1} \bmod \sigma(p^e)$. Also, $r=1$ if and only if $\sigma(p^e)$ divides $\sigma_k(p^e)$. Thus for $k$ coprime to $\tau(n)$, we have $\sigma(n)$ divides $\sigma_k(n)$. The relation also suggests that for a given $n$ the sequence $\sigma_k(n)\bmod \sigma(n)$ is periodic in $k$ with a period dividing $L$, the least common multiple of ($1+$ each exponent) in the prime factorization of $n$. Edit 2016.01.04: Once can show a nonreduced representation $\sigma_k(n) = a_k\sigma(n)/b_k$ where the $b_k$ are integers not necessarily coprime to the integers $a_k$ or to $a_k\sigma(n)$, with the property that the $b_k$ are bounded and periodic with period $L$. This is not enough to show $\sigma_k(n) \bmod \sigma(n)$ is periodic with small period, unfortunately. End Edit 2016.01.04.

If now $n$ is multiperfect (so $n$ divides $\sigma(n)$) we have $n$ divides $\sigma_k(n)$ for $k$ coprime to $\tau(n)$. In particular if $\tau(n)$ is a power of $2$, then $n$ divides $\sigma_k(n)$ for all odd $k \gt 0$.

It is still possible that $n$ can divide $\sigma_k(n)$ for $k$ not coprime to $\tau(n)$. However if $L$ is not prime, it seems likely that there will be more than one nonzero value of $\sigma_k(n) \bmod \sigma(n)$. If this is so, it would be one ingredient in a proof that 6 is the unique number having the titled properties, the other ingredient being that 6 is the only nontrivial multiperfect number with $L$ a prime.

Edit 2016.01.10: I botched an earlier edit which claimed that 6 is the only known multiperfect number $n$ which satisfies $\sigma_2(n) \bmod n = 2$. It is true, but the analysis had some flaws. However, one expects multiperfect numbers other than 1 and 6 to be a multiple of 4; when $n$ satisfies $\sigma(n) \bmod n = 0$ and $\sigma_2(n) \bmod n = 2$, and in addition $n \bmod 4 = 0$, then all odd prime factors of $n$ except one must occur to an even multiplicity, and the remaining odd prime factor must occur to a multiplicity of 1 mod 4 and must be a prime that is 3 mod 4. While simple, these observations say a lot about $n$ and suggest that any numbers satisfying the title congruences are rare indeed, perhaps more so than odd multiperfect numbers. End Edit 2016.01.10

Gerhard "Mea Culpa, Mea Maxima Culpa" Paseman, 2016.01.03

• I don't know why the downvote to my question – zeraoulia rafik Jan 3 '16 at 21:44
• I don't know why someone downvoted my answer. It turns out that 6 is the only multiperfect number with L a prime. So now it boils down to showing multiple nonzero values mod n for composite L. Gerhard "Will Accept Downvotes With(in) Reason" Paseman, 2016.01.03 – Gerhard Paseman Jan 3 '16 at 22:55
• The key idea in showing $n=6$ is the only multiperfect number with $L$ prime is to show all prime factors of $\sigma(n)$ and thus of $n$ are 0 or 1 mod $L$ all with multiplicity $L-1$, and thus $(L/(L-1))^{m+1} \geq \sigma(n)/n \geq L$, where $m$ is the number of distinct prime factors of $n$ that are 1 mod $L$. This quickly leads to $L=2$, and then showing the largest prime factor of $n$ must be at most one more than the second largest prime factor. Gerhard "It's Nice When Things Simplify" Paseman, 2016.01.04 – Gerhard Paseman Jan 4 '16 at 18:46
• I should say prime factors of $n$ have the same multiplicity. One has $\sigma(n)/n \geq L^{m+1-L}$ also. Gerhard "Dislikes Running Out Of Time" Paseman, 2016.01.04 – Gerhard Paseman Jan 4 '16 at 18:55
• 2+1 divides 2^3 + 1, 3+ 1 divides 3^5 + 1, 7+1 divides 7^11 + 1. So if e=1, p+1 divides p^(2k+1) + 1. Gerhard "Sometimes Confuses Odd And Even" Paseman, 2016.01.04 – Gerhard Paseman Jan 5 '16 at 0:37