Suppose $P < Q$ and $PQ=N$, where both $P$ and $Q$ are prime (including $2$). What's the distance $P$ from $\sqrt{N}$?
2026-02-23 06:20:58.1771827658
RSA semiprime distance between $P$ and $\sqrt{N}$.
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Claim. If we have $PQ = N$ and $P \le Q \le 4P$ for some positive integers (or positive real numbers, or primes) $P$ and $Q$, then it must be the case that $\sqrt N - P \le P$: the furthest $P$ can ever be from $\sqrt N$ is $P$.
Proof. From $Q \le 4P$, multiply by $P$ on both sides to get $PQ \le 4P^2$, or $N \le 4P^2$. Both sides are positive, so we can take the square root to get $\sqrt N \le 2P$. Subtracting $P$ from both sides, we get $\sqrt N - P \le P$.
Notes: