Why does setting the condition to x*x < n not work in finding prime factors?

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Setting the condition to x*x < n fails in finding prime factors because the value of n changes during the loop, impacting the exit condition. When a factor is found, n is divided by that factor, potentially making it smaller than x, which can lead to incorrect results. The correct approach is to use x < n, as this allows the loop to continue until all factors are found. The final output should be the last valid factor found, which is 6857 for the number 600851475143. Therefore, maintaining the original condition ensures accurate factorization.
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I've just started with project Euler. The problem I just finished is phrased as follows:

"The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?"

The method i used was trial division. Here is my code in C:
C:
#include<stdio.h>
int main()
{
     long long n = 600851475143;
     long long x;
     long long factor;

     for(x=2; x<n; x++)
     {
          while( n % x == 0 )
          {
                n =  n / x; 
                factor = x;
           }
     }

     printf("\n%lld\n", x);
     return 0;
}

Although this did solve the problem, (answer is 6572) I have read that you only need to check for factors upto the square root of n. I tried doing this by setting the condition x*x<n instead of x<n. This however, does not work and prints a value of 1472. Can someone explain why?
 
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Your code is wrong. Neither 6572 nor 1472 are factors of 600851475143. Notice that the variable factor is never used. I suggest getting it working on smaller numbers first. Also, think about what happens if you find a factor x, and this leads to a new value of n=n/x which is smaller than x.
 
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My apologies, i meant 6857. My program generated the correct answer. Admittedly I'm not too sure why and i don't know why it doesn't work when i try x*x<n. Also, i meant to print factor, not x.
 
CrazyNeutrino said:
My apologies, i meant 6857. My program generated the correct answer. Admittedly I'm not too sure why and i don't know why it doesn't work when i try x*x<n. Also, i meant to print factor, not x.

Does your original code (with x<n) work when you print factor?

Let me ask another question. Do you realize that when you update the value of n inside the for loop, that this impacts the test (x<n or x*x<n) which is done to decide when to exit the loop?
 
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CrazyNeutrino said:
Although this did solve the problem, (answer is 6572) I have read that you only need to check for factors up to the square root of n. I tried doing this by setting the condition x*x<n instead of x<n. This however, does not work and prints a value of 1472. Can someone explain why?
600851475143 = 71×839×1471×6857.

During the loop the values for n are:

600851475143/17 = 8462696833
8462696833/839 = 10086647
10086647/1471 = 6857
6857/6857 = 1

So the check for x < n is the correct check to use in this case. If n wasn't being divided each time a factor was found, then the check for x*x < n would work.
 
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