Observation on Giuga number

Question

•Define $S(a,m)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$

•Define $$f(m)=\sum_{S(a-1,m)\equiv -1\pmod a}1$$

Question

1) how to prove, For $m$ is odd, $f(m)=2$ and $a\in\{1,2\}$?

2) what is closed form to calculate $f(m)$ when $m$ is even?

Example $f(2)=3$ and $a\in\{1,2,3\}$

Edit:

I observed $a$ only belongs to $1$, prime number and Giuga number. e.g. $f(4)=5$ and $a\in\{1,2,3,5,30\}$ here $30$ is first Giuga number and $a$ also have it's prime factor.see, in $m=4,a$ have $2,3,5$ which prime factor of $30$. I also observed Carmichael number not belongs to $a$

Let $G$ denotes Giuga number and $p_1,p_2,...,p_n$ it's prime factors. e.g.$G=\prod p_i$

Statment (claim)

For some $m$, $p_i\in a \iff G\in a$.

Means $S(G-1,m)\equiv -1\pmod G\iff S(p_i-1,m)\equiv -1\pmod {p_i}$

Second edit: half solution for 2nd Q.

For $m$ is even $$S(p-1,m)\equiv \{0,-1\}\pmod p$$

$$S(p-1,m)\equiv -1\pmod p\implies p-1\mid m$$

$$\sum_{S(p-1,m)\equiv -1\pmod p}1= \sum_{p-1\mid m}1$$

Proof

By the above observation, if $a$ contains prime s.t prime factor of such giuga number then Giuga number also in set $a$.

So $f(m)=1+ \sum_{p-1\mid m}1+$ number of $G$ which generates by $p$.

Source code

n1= 1
o = 1
while n1 < 1000:
    m = 2

    num=n1
    sum_num = 0

    for i in range(1, num): 
        sum_num += i**(m)
    n2 = (sum_num)

    if((n2+1)%num == 0):
        print(n1,"diviasible")


    n1 += o

Code suggestions: don't use the letter o as a variable name. — Sean Roberson, Jan 11 '20 at 07:23
Please check, claim posted to MO: https://mathoverflow.net/q/350294/149083 — Pruthviraj, Jan 12 '20 at 20:13

score 2 · Answer 1 · answered Jan 11 '20 at 07:39

When $m$ is odd, we have: $$S(a-1,m)=1^m+2^m+\cdots+(a-1)^m$$ If $a-1$ is even: $$S(a-1,m)=(1^m+(a-1)^m)+(2^m+(a-2)^m)+ \cdots \bigg(\bigg(\frac{a-1}{2}\bigg)^m+\bigg(\frac{a+1}{2}\bigg)^m\bigg)$$ $$S(a-1,m)\equiv0 \pmod{a}$$ If $S(a-1,m) \equiv -1 \pmod{a} \implies a \mid (-1) \implies a=1$

Thus, in this case, $a=1$ is the only solution.

If $a-1$ is odd: $$S(a-1,m)=(1^m+(a-1)^m)+(2^m+(a-2)^m)+ \cdots \bigg(\bigg(\frac{a-2}{2}\bigg)^m+\bigg(\frac{a+2}{2}\bigg)^m\bigg)+\bigg(\frac{a}{2}\bigg)^m$$ $$S(a-1,m)\equiv \bigg(\frac{a}{2}\bigg)^m \pmod{a}$$ If $S(a-1,m) \equiv -1$, let $p \mid a$ for prime $p$. We cannot have $p \mid \frac{a}{2}$ since: $$\bigg(\frac{a}{2}\bigg)^m \equiv -1 \pmod{a}$$ Thus, we must have $\frac{a}{2}=1 \implies a=2$.

Thus, $a=2$ is the only solution in this case. This finally shows that when $m$ is odd, we have $f(m)=2$ where $a$ only takes the values $1$ and $2$.

Observation on Giuga number

1 Answers1