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Let $a,b \in \mathbb{N}$. Denote the set of prime numbers by $P$.

Here it was shown that the probability that $\gcd(a,b) \in P$ is $0.2749 \ldots$.

What is the probability that $\gcd(a,b) \in 2P$?

Any hints and comments are welcome!

Edit: In the above link it is also mentioned that the probability that $\gcd(a,b)=1$ is $0.6079\ldots$. Then the probability that $\gcd(a,b) \in \{1\} \cup P$ is $0.6079\ldots + 0.2749 \ldots = 0.8822 \ldots$. Thereore, the probability that $\gcd(a,b) \in 2P$ should be strictly less than $0.1178 \ldots$.

(Also notice that the probability that $\gcd(a,b) \in \{1\} \cup P \cup P^2 \cup P^3 \cup \cdots=\{1\} \cup_{i=1}^{\infty} P^i$ is $1$, we must have 'very small' probabilities that $\gcd(a,b) \in P^m$, for a fixed $m \geq 2$).

user237522
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1 Answers1

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Hint: For this to hold, $a$ and $b$ must both be even, and after dividing both by 2, their gcd must be prime. Can you use that to figure out the answer?

rogerl
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  • Thank you for the hints. So the probability is also $0.2749\ldots$? Since we can write $a=2pA$ and $b=2pB$, with $p$ prime and $\gcd(A,B)=1$, and consider $\frac{a}{2}$ and $\frac{b}{2}$ instead? We have, $\gcd(\frac{a}{2},\frac{b}{2})=\gcd(pA,pB)=p$. – user237522 Oct 17 '18 at 17:18
  • @user237522 Not exactly, as "uniformly chosen" $A$ and $B$ aren't necessarily even... – Carl Schildkraut Oct 17 '18 at 17:43
  • @CarlSchildkraut, thanks for the hint. However, I still do not know the answer.. – user237522 Oct 17 '18 at 17:56
  • You should look more carefully at the linked article. The answer there can tell you how (for fixed $p\in P$) the probability of $\gcd(a,b)=p$ relates to the probability of $\gcd(a,b)=2p$. – rogerl Oct 17 '18 at 18:19
  • @rogerl, thank you. (Could you please elaborate your answer?). – user237522 Oct 17 '18 at 20:11