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When reading the book Mersenne Numbers and Fermat Numbers, after proving that: for any positive integers m,n, it holds $\gcd(M_n,M_m)=1$ if and only if $\gcd(m,n)=1$, it says that this allows us to give one more proof of the infiniteness of the set of prime numbers:

Here is the proof: Consider the set $\{M_2, M_3,\cdots, M_{p_n},\cdots\}$,of all Mersenne numbers with prime indexes $p_1=2, p_2=3$, $\cdots$. For each $M_{p_i}$, denoted by $q_i$ the smallest prime divisor of $M_{p_i}$. In this way, we get a set $\{q_1,q_2, \cdots, q_i, \cdots\}$ of prime numbers: $q_1=3, q_2=7$, etc.

For $i\ne j$, it holds $p_i \ne p_j$, and hence $\gcd(p_i,p_j)=1$; so $\gcd(M_{p_i},M_{p_j})=1$, and $q_i\ne q_j$. Therefore, all elements of the set $\{q_1,q_2, \cdots, q_i, \cdots\}$ are different. If the set P of primes of finite say, $P=\{p_1,p_2,\cdots, p_k\}$, it should coincide with the set $\{q_1,q_2, \cdots, q_k\}$. But $p_1=2$, while all numbers $q_i$ are odd; a contradiction. It proves that the set P of primes is infinite.

I don't quite understand the sentence in bold (why P should coincide with the set containing $q_i$). Can anyone help me?

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    $\begingroup$ Well, more simply: consider the $M_p$ for all primes $p$. These are relatively prime so the least prime dividing each gives us an infinite list of primes. $\endgroup$
    – lulu
    Commented Jun 6, 2023 at 14:37

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If $P$ were the set of all primes and were assumed to be finite $P =: \{p_1, \dots, p_k\}$, then it would have to contain $\{q_1,\dots, q_k\}$, since they are prime numbers, but all the $q_i$s are different per what has been said, thus we would have equality of cardinals between $\{p_1,\dots, p_k\}$ and $\{q_1,\dots,q_k\}$ with an inclusion, thus we would have equality of the sets.

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