I read the statement in the title somewhere but I can't find any proof.

For a positive integer $n$, why can't there be 4 numbers $a, b, c, d$ ($b$ and $d$ are prime) for which

$a < \sqrt{n} < b$, where $a \cdot b = n$


$c < \sqrt{n} < d$, where $c \cdot d = n$

  • 4
    $\begingroup$ Use that if for different primes $p$ and $q$ you know $p|n$ and $q|n$ then you know $pq|n$. Can you now deduce a contradiction using inequalities? $\endgroup$ – Tintarn Aug 24 '15 at 22:09
  • $\begingroup$ I am not sure but $16 = 2^4$ has only prime factor (2) and $2<4=\sqrt{16}$ $\endgroup$ – GAVD Aug 25 '15 at 2:02
  • $\begingroup$ @GAVD it is a bit confusing.. It doesn't say that there is one, it says that there's at most one $\endgroup$ – Gil-Mor Aug 25 '15 at 7:19
  • $\begingroup$ @Gil-Mor: Ah thank you! $\endgroup$ – GAVD Aug 25 '15 at 7:49

If $p$ and $q$ are primes that are $>\sqrt{n}$, and $n$ is divisible by $p$ and $q$, then $n$ is divisible by $pq$. But $pq>n$, which is a contradiction.


Assume by contradiction that there exists a number $n\in\mathbb{N}$ with two prime factors $p,q>\sqrt{n}$:

  • $p$ and $q$ are prime factors of ${n}\implies\color\red{{p}\cdot{q}\leq{n}}$
  • $p,q>\sqrt{n}\implies{p}\cdot{q}>\sqrt{n}\cdot\sqrt{n}\implies\color\red{{p}\cdot{q}>{n}}$

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