It has already been known that if n be a even perfect number then n must have the form :$2^{p-1}(2^p-1)$ where $p,2^p-1$both are primes.

An odd perfect number does not fit since $2| n + 1$ and $n = 1$ is not perfect. For $p > 2$ we have \begin{eqnarray*} 2^{p - 1} \cdot \left( 2^p - 1 \right) - 1 & \equiv & ( - 1 )^{p - 1} \cdot \left( ( - 1 )^p - 1 \right) - 1\\ & \equiv & 1 \cdot \left( - 2 \right) - 1\\ 2^{p - 1} \cdot ( 2^p - 1 ) - 1 & \equiv & 0 ( \mbox { mod } 3 ) \end{eqnarray*} But $2^{p - 1} \cdot ( 2^p - 1 ) - 1 \geq 27$ is supposed prime. Impossible. The case $p=2$ gives the desired solution, which is $n=6$ [corrected a slight mistake]

Quite same aprroach is the following : The number n will have the form $2^{k-1}(2^k-1)$ where $2^k-1$ is a prime number .from the problem $n-1,n+1$ are primes so $3|n$ .So $3|(2^{k-1}(2^k-1))$ so $2^k-1=3$ so k=2 and n=6 is the unique solution

If n=2k+1 then one of n-1 or n+1 isnt a prime number.So 2 divides n.If 3 doesnt divide n then one of n-1 or n+1 isnt a prime number.So 3 divides n.Hence 6 divides n.If n=6 we get a solution.Let n=6r(r>=2).Then 2n=12r=3r+2r+r+2+3+6+1+n.Hence2+3+6+1=0.Contradiction.If some of the divisors match(for example 3r=6)we can check that there is no solution.

Hawk Tiger wrote: It has already been known that if n be a even perfect number then n must have the form :$2^{p-1}(2^p-1)$ where $p,2^p-1$both are primes. It's the Euler's theory

Hawk Tiger wrote: It has already been known that if n be a even perfect number then n must have the form :$2^{p-1}(2^p-1)$ where $p,2^p-1$both are primes. Since this solves 90% the problem, you have to prove that.

Yes ,you are right,freemind.I just think the others should have the chance to solve this one by themselves,So I didn't post the solution.

Theorem: If $a$ is a perfect even number then there exist $n$ such that $a=2^{n-1}(2^n-1)$ Proof : Because $a$ is even it will have the form $a=2^{n-1}b$ where $b$ is an odd number .We have $s(b)=\frac{2^nb}{s(2^{n-1}}$ where $s(k)$ is the sum of divisors of $k$ . So $2^n-1|b$ and $s(b)=b+\frac{b}{2^n-1}$ . From this $\frac{b}{2^n-1}=1$ and the conclusion follows

Obviously n=6k, k is a positive integer.If k is bigger than 1 then the number 1+k+2k+3k is bigger than 6k=n and 1,k,2k,3k are divisors of n so it contradicts to n is perfect so k=1 and n=6.It is clear that 6 is perfect

Which is the correct and small solution? I dont understand

6 is the only solution. If the number is bigger than 6, it's of form 6k, because primes are only 6k+1 and 6k-1. But 6k can't be perfect because 1,k,2k,3k,6k are its distinct divisors of 6k (k>1) and their sum is bigger than 12k, so k=1 Is this what you wanted?

many thank bugi this is very simle solution

The first perfect number is $n = 6,$ which satisfies the conditions, so assume that $n > 6.$ We claim that no perfect number $n > 6$ has both $n-1$ and $n+1$ prime. Suppose otherwise; then, since $n-1$ and $n+1$ cannot be multiples of $3,$ $n$ must be a multiple of $3.$ Similarly, $n$ must be even since $n-1$ and $n+1$ must both be odd. Then, $n$ is a multiple of $6$. Since $n > 6,$ the numbers $n, \dfrac{n}{2}, \dfrac{n}{3}, \dfrac{n}{6}$, and $1$ are pairwise distinct and divisors of $n,$ so the sum of the divisors of $n$ is at least \[n + \dfrac{n}{2} + \dfrac{n}{3} + \dfrac{n}{6} + 1 = 2n+1.\] Thus, the sum of the divisors of $n$ is strictly greater than $2n,$ so $n$ cannot be perfect, a contradiction. We conclude that $n = \boxed{6}$ is the only such integer. $\square$ (Similar to Bugi's.)