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I know Euclid's clever way of proving that there are infinitely many primes but while I was studying, I also saw a sentence on the book: "Euler-phi function can also be used to show that there are infinitely many primes." but could not figure out how. Could you please help me to understand it?

Regards

Amadeus
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Assume that there are only a finite number of primes say $p_1,p_2,\ldots,p_k$. Look at the product of these finite primes i.e. $$m = p_1 p_2 \cdots p_k$$ Now consider any number $n > 1$. Since there are only finite primes, one of the $p_j$'s must divide $n$. Hence, $\gcd(m,n) > 1$. Hence, $\phi(m) = 1$.

Can you now finish it off?

  • I guess it must go on like this: we know that $\phi(m) = 1$ however according to the multiplicative property of the phi function, $\phi(m) = (p_1-1)(p_2-1)...*(p_k-1)$ which is not clearly equal to 1. However I could not exactly understand why $\phi(m) = 1$. Maybe I should give it some time. = ) – Amadeus Nov 30 '12 at 02:27
  • I guess I understood why $\phi(m) = 1$. Any selection of (m,n) will not be relatively prime only (m,1) will be relatively prime hence $\phi(m) = 1$. That is brilliant. Thank you so much. – Amadeus Nov 30 '12 at 02:30
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    @Zxy Precisely. $\phi(m) = #{n \leq m: \gcd(m,n) = 1}$ –  Nov 30 '12 at 02:32
  • @Zyx $\rm:m \ge 2:\Rightarrow:1\ne m!-!1:$ are both coprime to $\rm:m:$ so $\rm:\phi(m)\ge 2.:$ This is just the $\rm:m!-!1:$ (vs. $\rm:m!+!1):$ form of Euclid's proof. – Bill Dubuque Nov 30 '12 at 03:39