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My proof is:

Let $N=pq$. If $p=k\sqrt{N}$ where $k\in\mathbb{Q}^+$, then $q=\dfrac{\sqrt{N}}{k}$.

If $k=1$, clearly $\sqrt{N}$ is a factor of $N$.

If $k<1$, then $p<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

If $k>1$, then $q<\sqrt{N}$, so there must be at least one factor of $N$ less than $\sqrt{N}$.

Therefore, there always exists at leastone number less than or equal to $\sqrt{N}$ that divides $N$. So, if there is no such number, $N$ is definitely prime.

Where did I go wrong?

ultralegend5385
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1 Answers1

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$1$ is "at least one number less than or equal to $\sqrt{N}$ that divides $N$".

Eric Towers
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  • Can we simply add the restriction $p,q\neq 1$ and then will the proof be okay? – ultralegend5385 Jan 11 '21 at 08:31
  • @ultralegend5385 : Not precisely. First, you never actually say $p$ and $q$ are integers, which you should. Second, if $N$ is not a square, then $k \sqrt{N}$ is irrational for every $k \neq 0$, so $k \sqrt{N}$ cannot be the integer $p$ (and similarly $q$, since division by nonzero rational numbers is the same as multiplication by rational numbers). That is, $p$ and $q$ cannot be of the form you specify. – Eric Towers Jan 11 '21 at 08:37
  • Fine! I got it. – ultralegend5385 Jan 11 '21 at 08:41