
Congruence problem?
Suppose that the positive integer n has the property that is divisible by 3.
if , where and are distinct odd primes with , then p = 3 and ).
if then
working modulo 3, p can take any of the values 0, 1, 2 but q can take only values 1 or 2 since is prime.
Now, I can understand why p can only take values 0, 1, 2, but why can q only take values 1 or 2? q > 3 is prime? I don't understand what is meant by this. Obviously and is not prime for all q, and if q is defined as a odd prime this is just a pointless statement, so what does it mean and why does it put restrictions on what values q can take modulo 3?

Re: Congruence problem?
reading this I'm not sure what you're trying to prove and I'm not convinced you understand what $\sigma(n)$ means.
Could you post the original problem?

Re: Congruence problem?
If a number is congruent to 0 modulo 3, that means it is divisible by 3. But, is a prime greater than 3, so it is not divisible by 3. Since it is prime, it is only divisible by 1 and itself.

Re: Congruence problem?
I think I see this now. Is it that as and is the smallest odd prime, will always have to be larger than to satisfy ? And seeing as is the only prime modulo 0 to 3 it only leaves the other two least positive residuals 1 and 2 as possibilities.
is the sum of divisors of .

Re: Congruence problem?
Hi,
Here's your problem as I understand it:
Let n=pq be the product of odd primes p and q with p<q. Assume 3 divides the sum $s=n+\sigma(n)=pq+(p+1)(q+1)$. Then p is 3 and q is congruent to 5 mod 6.
Hints on the proof:
First suppose 3 does not divide n; i.e. neither p nor q is 3. Then both p and q are nonzero mod 3. Consider the possibilities for p mod 3 and q mod 3.
1. Both are 1 mod 3. Then s is $1+2\cdot2=2$ mod 3, a contradiction.
2. Similarly for both 2 mod 3.
3. Similarly for one is 1 mod 3 and the other is 2 mod 3.
So 3 divides n and thus p=3.
Once p=3, you should be able to quickly get that q is congruent to 5 mod 6 by first showing that q is 2 mod 3 and then use the fact that q is odd.