this is just perfect!!!!!!!!!

let us consider the only two cases possible one by one let N= $n^{n}$ +1 CASE 1:- let n be odd.since N is an even perfect number,N must of the form N= $2^{m-1}$ ( $2^{m}$ -1),(by euclid's theorem),where $2^{m-1}$ is a prime. clearly,N can be factored as N = $n^{n}$ +1 = (n+1)r,where r= $n^{n-1}$ - $n^{n-2}$ + ... -n+1.we now claim that (n+1,r)=1.to show this,notice that since n is odd ,r is odd and n+1 is even.let n+1= $2^{s}$ t,where t is an odd positive integer.then N= $2^{s}$ tr,where both t and r are odd.since N is an even perfect number,this possible only if t=1;so n+1= $2^{s}$ and hence (n+1,r)=1(notice that if r=1,then N= $n^{n}$ +1=n+1;so n=1.then N=2 which is not a perfect number) Since N= $2^{m-1}$ ( $2^{m}$ -1)=(n+1)r= $2^{s}$ r ,where $2^{m-1}$ is a prime and r is odd, $2^{s}$ = $2^{m-1}$ = $n+1$ and r= ( $2^{m}$ -1)=2. $2^{m-1}$ -1 = 2(n+1)-1=2n+1 Therefore, N= $n^{n}$ +1= $(n+1)(2n+1)$ = $2n^{2}$ +3n+1 this yields $n^{n}$ = $2n^{2}$ +3n
$n^{n-1}$ = $2n+3$ since n is an integer this equation has an unique solution 3 it can be verified algebraically. then N= $n^{n}$ +1=28 thus 28 is the only even perfect number!

further, i claim that there are no odd perfect numbers of the form $n^{n}$ +1. thus our answer is 28. if anyone can furnish us with a proof of the non existence of an odd perfect number of the mentioned form it will delightful if not,i will post it at a future date