# Is there a Non-prime Number that is Divisible only by Numbers Greater than its Square Root? [duplicate]

Possible Duplicate:
Calculating prime numbers

The question is in the title. Is there a number that is divisible only by numbers greater than its square root? If not, why? I need this because it can speed up a calculation algorithm significantly if the answer is no.

• No. See here. If $n$ has a divisor $d \geq \sqrt{n}$ then $1/d \leq 1/\sqrt{n}$ and thus $n/d \leq n/\sqrt{n} = \sqrt{n}$.
– t.b.
Jun 19, 2011 at 18:40
• If this is closed as a duplicate of the other question, its title should probably be changed — "Calculating prime numbers" is not a very illuminating title or a good search target. Jun 19, 2011 at 18:45
• Thank you. I think that the reason why this might have come up before is that it can significantly reduce calculation time in any prime-finding algorithm. Jun 19, 2011 at 18:47
• Note:This fact increased my program's speed exponentially Jun 19, 2011 at 18:57
• Exponentially? What kind of program did you have before? Jun 19, 2011 at 19:21

If $a$ is not a prime, $b | a$ and $\sqrt a < b < a$, then $\frac a b | a$ and $1 < \frac a b < \sqrt a$. So the answer is obviously no.
No. Suppose that $n$ is positive. If $a > \sqrt n$ and $b > \sqrt n$ then $a b > (\sqrt n)^{2} = n$. Thus no positive number $n$ can be the product of two numbers, each of which is greater than the square root of $n$.
No. If $n=ab$ and $a>\sqrt{n}$, or in other words $a>\sqrt{ab}$, then by squaring both sides we have $a^2>ab$. Multiplying both sides by $\frac{b}{a}$, we have $ab>b^2$. Taking the square root of both sides, we have $\sqrt{ab}>b$, i.e. $\sqrt{n}>b$. Thus, every composite number is divisible by numbers less than its square root. In fact, the divisors of $n$ greater than $\sqrt{n}$ and the divisors of $n$ less than $\sqrt{n}$ are in bijection, by sending $d\mapsto \frac{n}{d}$.
• I.e. the integer points of the hyperbola $\:xy = n\:$ are preserved by the reflection $\:(x,y)\mapsto (y,x)\:.$ Jun 19, 2011 at 22:53