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This question already has an answer here:

Provide an infinite sequence of natural numbers $x_1,x_2,x_3,\ldots$ such that I'm not sure if I'm on the right track with that. any help would be great

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marked as duplicate by Community Mar 15 '17 at 5:19

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Your answer is certainly correct: if you choose $x_n$ to be $p_n^2$ (the square of the $n$th prime number), then

  • none of the numbers $x_n$ are prime numbers $\checkmark$, and
  • any two $x_n$ and $x_m$ are coprime to each other $\checkmark$

The latter part is true because

  • $\gcd(p,q)=1$ for any distinct prime numbers $p$ and $q$
  • $\gcd(ab,c)=\gcd(a,c)$ if $\gcd(b,c)=1$, for any integers $a,b,c$

and therefore $$\gcd(x_n,x_m)=\gcd(p_n^2,p_m^2)=\gcd(p_n,p_m^2)=\gcd(p_m^2,p_n)=\gcd(p_m,p_n)=1$$ for any two prime numbers $p_n$, $p_m$.

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  • $\begingroup$ ahh thanks ! how would you write it so theres an infinite sequence of natural numbers ? $\endgroup$ – rprogramr Mar 14 '17 at 8:58
  • $\begingroup$ @rprogramr: Not sure what you're looking for; I would say that the sentence $$\text{“Define $x_n$ to be $p_n^2$, the square of the $n$th prime number”}$$ gives an infinite sequence of natural numbers, since there are infinitely many prime numbers. $\endgroup$ – Zev Chonoles Mar 14 '17 at 9:03
  • $\begingroup$ alright I see thank you for the feedback ! :) $\endgroup$ – rprogramr Mar 14 '17 at 9:04

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