code not working

[gelatine]

here is what my code should do:

A perfect number is a number for which the sum of its proper divisors is exactly 
equal to the number. For example, the sum of the proper divisors of 28 would be 1 
+ 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is
 called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
 number that can be written as the sum of two abundant numbers is 24. By 
mathematical analysis, it can be shown that all integers greater than 28123 can be
 written as the sum of two abundant numbers. However, this upper limit cannot be 
reduced any further by analysis even though it is known that the greatest number 
that cannot be expressed as the sum of two abundant numbers is less than this 
limit.

Find the sum of all the positive integers which cannot be written as the sum of two
 abundant numbers.

(or take a look at http://projecteuler.net/problem=23
here is my code:

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <fstream>
#include <string>
#include <time.h>
using namespace std;
int abundant[28124];
int size;

int d(int n) {
int divisors=0;
	for (int i=1;i<=sqrt(double(n));i++) {
		if (n%i==0) { 
			divisors+=i;
			divisors+=n/i;
		}
	}
	return divisors-n;
}

bool summable(int x){
	for (int i=0;i<size && abundant[i]<=x;i++) {
		for (int j=0;j<size && abundant[j]<=x;j++) {
			if (abundant[i]+abundant[j]==x) {
			return true;
			}
		}
	}
	return false;
}

int main() {
	clock_t start, end;
	start = clock();
	int sum=0;
	int count=0;
	int count2=0;
	for (int i=0;i<28124;i++) {
		if (d(i)>i) {
			abundant[count]=i;
			count++;
			count2++;
		}
	}
	size=count2;
	cout<<size<<endl;
	for (int i=0;i<28124;i++) {
		if (!summable(i)) {
			sum=sum+i;	
		}
	}
	cout<<sum<<endl;
	end= clock();
	printf("\nTook %f seconds\n", (double)(end-start)/CLOCKS_PER_SEC);
	system("pause");
	return 0;
}

the problem is that the answer i get from the program is wrong. with this program i get around 3900000 but the answer should be around 4100000
can someone tell me what i am doing wrong ?

[helios]

d() is incorrect. If the number is a perfect square, its square root gets added twice into the result.

[gelatine]

thank you. it is solved now.
for someone who's interested i added

if (floor(sqrt(double(n)))==sqrt(double(n))) {
		divisors=divisors-sqrt(double(n));
	}

before the return in d()