Weird situation in prime number checking code

Question

When I was solving a problem for Project Euler, it asked me to sum up all the primes below 2 million. Here is my code:

#include<stdio.h>
#include<math.h>
int isPrime(int);
int main() {
    long long int sum = 0;
    int i; // index
    for(i = 2 ; i < 2000000 ; i++) {
        if(isPrime(i)) {
            sum += i;
        }
    }
    printf("%lli\n", sum);
}

int isPrime(int num) {
    int i; // index
    int sq = sqrt(num);
    for(i = 2 ; i <= sq ; i++) {
        if(num % i == 0) {
            return 0;
        }
    }
    return 1;
}

This code leads to the correct answer, 142913828922. But when I change the for loop in isPrime() to:

for(i = 2; i <= sq+1; i++)   // or even sq+2, sq+3, etc.

It leads to incorrect results such as 142913828920 and 142913828917, etc.

Why does it go wrong? Theoretically, it doesn't change the number isPrime() sends to main(), does it?

Answer 1

if you change the loop to

for(i = 2 ; i <= sq+1 ; i++)

then 2 is no longer considered to be prime, because you test if 2 % 2 == 0.

Similar for larger numbers you add, more and more primes will not be detected as such.

Answer 2

Considering you changed the sum from 142913828922 to 142913828920, then the difference is 2, which means you're interpreting 2 as not prime. Changing sq to sq+1 should achieve that difference. Changing it to sq+2 would end up making 3 not prime.

((int)sqrt(2))+1 == 2
((int)sqrt(3))+2 == 3

and so on.