On the sum of divisors

For discussing Olympiad Level Number Theory problems
User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

On the sum of divisors

Unread post by Masum » Thu Dec 09, 2010 4:21 am

(Masum Billal): A number has $p$ divisiors where $p$ is a prime.Decide if it has some divisors (excluding itself) such that their sum is equal to the initial number.
One one thing is neutral in the universe, that is $0$.

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: On the sum of divisors

Unread post by Moon » Thu Dec 09, 2010 1:15 pm

Here goes the hint (hidden below)
If $n$ is the number then $n=q^{2^m\cdot k}$ where $q$ is a prime and $p=2^m\cdot k+1$
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

Re: On the sum of divisors

Unread post by Masum » Thu Dec 09, 2010 6:38 pm

Post your full solution.
Here is my solution:
Since $p$ prime,ofcourse,$n=q^{p-1}$ for $q$ prime.
First solution:The divisors are $1,q,q^2,...,q^{p-2}$ and there sum is $\frac {q^{p-1}-1} {q-1}$ which is less than $q^{p-1}$.So the sum of some divisors can't be equal to the number.
Second solution:Let $a_0,a_1,...,a_{p-2}$ be a permutation of $0,1,..,p-2$.So if $q^{p-1}=\sum q^{a_i}$ for some $0\le i\le p-2$,then let $a$ be the smallest of $a_i$.And we find contradiction after dividing by $q^a$
One one thing is neutral in the universe, that is $0$.

Post Reply