Sum of prime generating numbers. I was trying to solve a project euler problem which goes like : 

Find the sum of all positive integers $n$ not exceeding $100,000,000$
  such that for every divisor $d$ of $n$, $d+\frac nd$ is prime.

So this is what I did.
if $d + \frac nd$ is prime and $1$ is always a divisor of $n$ thus for $d=1$, this should be true. Thus we'll get $1+n$ should be prime. Meaning $n$ is even.
Now if $n$ is even then $2$ becomes a divisor of $n$. Thus $\left(2+\frac n2\right)$ should also be prime which means $\frac n2$  should be odd and so we get $n$ should not be a multiple of $4$. 
So from above what we have is $n+1$ is prime and $n$ is not a multiple of $4$.
I couldn't find any more constraints on $n$.
How do I proceed further? 
I have a feeling that these are all the constraints I need on $n$ but I'm not able to prove this.  
Edit: As peter in comments suggested $n$ also needs to be square-free. So we have another condition on $n$.
 A: The following PARI/GP programs determines the number of positive integers $n$ upto $10^8$ satisfying the condition and the sum of those :
? z=0;su=0;forprime(p=2,10^8+1,n=p-1;if(issquarefree(n)==1,gef=1;fordiv(n,d,if(gef==1,if(ispseudoprime(d+n/d)==0,gef=0)));if(gef==1,su=su+n;z=z+1)));print(z,"  ",su)
39627  1739023853137

The results for the number of prime factors of $n$ :


*

*0 factors : 1/1

*1 factors : 1/2

*2 factors : 30301 / 1358128392798

*3 factors : 9009 / 370999149930

*4 factors : 312 / 9892176864

*5 factors : 3 / 4133542

A: Here's what I've got:


*

*You only really need to care about divisors below the square root of $n$ because $d_1d_2=n$ implies that $d_1+{n\over d_1}=d_1+d_2=d_2+d_1=d_2+{n\over d_2}$ and the lower is always going to be below the square root. That results in divisors no larger than $10^4$ needing to be checked.

*By similar logic, you can avoid numbers that aren't squarefree. Because, you can split the power over the divisors, leading to both addends having a common factor, which factoring out would find.

*by the product  of squares is a square, you really only need to check for divisibility by prime squares ( basically what a squarefree check is, just thought I'd mention it)

*The properties of being 1 less than a prime, and squareefree means that prime can only be the prime 2; or 3 mod 4.  Not being 1 mod 9 ( the next square) means not being 1 mod 18, which then has only 1 case mod 36 that works due to 1 mod 36 being covered by 1 mod 4, we can also eliminate primes that are 19 mod 36 then. 

*we can then eliminate primes that are 51 mod 100 etc.   
