Exercise 2.3

Proof. Let $\{p_1,\cdots ,p_t\}$ the finite set of primes,with $p_1<p_2<\cdots <p_t$, and $n=p_1\cdots p_t$. By definition, $\varphi (n)$ is the number of integers $k,1\le k\le n$, such that $k\wedge n=1$. From the existence of decomposition in primes, if $k\ge 1$, $k=p_1^{k_1}\cdots p_t^{k_t}$, where $k_i\ge 0,i=1,\dots ,t$. So $k\wedge n=1$ if and only if $k=1$. Thus $\varphi (n)=1$ The formula for $\varphi (n)$ gives $\varphi (n)=(p_1-1)\cdots (p_t-1)=1$. As $p_i\ge 2$, this equation implies that $p_1=p_2=\cdots =p_t=2$, so $t=1$, and the only prime number is 2. But 3 is also a prime number : this is a contradiction.

Conclusion : there are infinitely many prime numbers. □