why only square root approach to check number is prime [closed]
Solution 1:
If $a\cdot b=N$ where $1<a\le b<N$
$N=ab\ge a^2\iff a^2\le N\implies a\le\sqrt N$
Solution 2:
A more intuitive explanation of the same thing. For the same reasons.
The square root of 100 is 10. Let's say a x b = 100, for various pairs of a and b.
If a == b, then they are equal, and are the square root of 100, exactly. Which is 10.
If one of them is less than 10, the other has to be greater. For example, 5 x 20 == 100. One is greater than 10, the other is less than 10.
Thinking about a x b, if one of them goes down, the other must get bigger to compensate, so the product stays at 100. They pivot around the square root.
The square root of 101 is about 10.049875621. So if you're testing the number 101 for primality, you only need to try the integers up through 10, including 10. But 8, 9, and 10 are not themselves prime, so you only have to test up through 7, which is prime.
Because if there's a pair of factors with one of the numbers bigger than 10, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor of 101.
If you're testing 121, the square root is 11. You have to test the prime integers 1 through 11 (inclusive) to see if it goes in evenly. 11 goes in 11 times, so 121 is not prime. If you had stopped at 10, and not tested 11, you would have missed 11.
You have to test every prime integer greater than 2, but less than or equal to the square root, assuming you are only testing odd numbers.
Solution 3:
If n is a sum of 2 addends, one addend must be $\leq n/2$
If n is a sum of 3 addends, one addend must be $\leq n/3$
If n is a sum of 4 addends, one addend must be $\leq n/4$
etc...
If n is a product of 2 factors, one factor must be $\leq \sqrt{n}$
If n is a product of 3 factors, one factor must be $\leq \sqrt[3]{n}$
If n is a product of 4 factors, one factor must be $\leq\sqrt[4]{n}$
etc...
Since having 2 factors is sufficient to be non-prime, $\sqrt{n}$ is used to bound the test.
Solution 4:
Because it is true that every composite number $N$ has a prime factor at most equal to $\sqrt N$ but it is not true that every composite number has a prime factor at most equal to $\sqrt[k] N$ for $k\ge 3$. For instance, $N=p^2$, where $p$ is prime, has no prime factor at most equal to $\sqrt[k] N$ for any $k\ge 3$.