What's the relation between factors of a number and its square root?

Question

For instance, if the square root of a number $N$ is an integer, $N$ is a square number. But for instance $\sqrt{80} = 8.944...$, the fractional part is close to an integer, and indeed $81$ is a square number, with the factor $9$. The factors of $80$ are $10$ and $8$, so they're still very close to $9$. Is this coincidental (aside from the fact that $80 = 9^2-1^2$), or is there any way to approximate the integer factors of a number, given the square root of it?

Answer 1

Primes don't have any (non trivial) integer factors, so there is nothing to approximate. So I guess the answer to your question is no.

Answer 2

Obviously we can start with an arbitrary natural number $N$, take its square root, and round this value to the nearest integer $M$. We can then define the (small) remainder $R$ as the difference between $N$ and $M^2$. This way we arrive at the equation:

$$N = M^2 + R$$

Now the question by the OP is whether this helps us to find the integer factors of $N$. In general the answer must be no. However, there are exceptions, such as $R = -P^2$. Then we find

$$N = M^2 - P^2 = (M-P) (M+P)$$

We see that in this case there are indeed two factors in $N$ which are both close to $M$. The behaviour observed by the OP for $N=80, M=9, R=-1, P=1$ is therefore not "coincidental".