Elementary number theory sum of divisors

Question

Let the sum of the divisors of a number $N$ be equal to $s$(excluding N itself) then show that if $s=N$ then show that N is a perfect number.

I tried to use the basic formula for sum of divisors but that surely didn't work out anything. Can anybody please give me some hints as to how to prove the given statement.

Answer 1

$Hint$ : By the definition of a perfect number ; "A positive integer N is said to be perfect if N is equal the sum of all its positive divisors excluding N itself . " Also the sum of divisors of a perfect number is twice the number .

Answer 2

$$N=P_1^{\alpha_1}\times\,P_2^{\alpha_2}\times\,...\times\,P_k^{\alpha_k}$$

$$\sigma(N)=\sum\limits_{d|n,d\ne\,N}{d}=\frac{1}{2}\prod\limits_{i=1}^{k}{\sigma (P_{i}^{{{\alpha }_{i}}})=\frac{1}{2}\prod\limits_{i=1}^{k}{\frac{P_{i}^{{{\alpha }_{i}}+1}-1}{{{P}_{i}}-1}}}$$

Answer 3

Let $\sigma(x)$ be the sum of all the divisors of the positive integer $x$. For example, $$\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.$$

If we let $s(x) = \sigma(x) - x$, then $s(x)$ is the sum of the aliquot parts of $x$.

Suppose that $s(N) = N$. Then we have $$\sigma(N) - N = N$$ which implies that $$\sigma(N) = 2N.$$ This last equation is the definition of a perfect number, in terms of the classical sum-of-divisors function $\sigma$.