I am having difficulty in understanding the proof of Behmann of Infinitude of primes. Can someone please explain the last part 'The proof is concluded by noticing….' which is in page $178$?
Any help would be appreciated. Thanks in advance.
Behmann’s proof of Infinitude of primes.
algebraic-number-theoryanalytic-number-theoryelementary-number-theorynumber theoryprime numbers
Related Solutions
Let me first prove that the number of primes is infinite. This can be achieved using the identity $$ x = \sum_{d \ge 1} \mu(d) \frac{x^d}{1-x^d} , $$ where $\mu$ denotes the Moebius function.
If there are only finitely many prime numbers $p_1$, $\ldots$ , $p_n$, then no integer larger than $N = p_1 \cdots p_n$ can be squarefree, hence $\mu(d) = 0$ for all $d > N$ but $\mu(N) \ne 0$. But then the function on the right has a pole in $x = e^{2\pi i/N}$, whereas the polynomial on the left is entire. This is a contradiction. I am sure that there are more clever ways of obtaining such a contradiction.
The identity used for proving that there are infinitely many primes $q \equiv 3 \bmod 4$ is $$ \sum_d x^d = \sum_m \mu(m) \frac{x^m}{1-x^{2m}}, $$ where the sum on the left is over all odd natural numbers $d$ not divisible by any prime $q \equiv 3 \bmod 4$, and the sum on the right is over all odd integers $m$ not divisible by any prime $p \equiv 1 \bmod 4$.
Assume that there are only finitely many primes $q \equiv 3 \bmod 4$. Then the sum on the right is finite, and the last nonzero term occurs when $m$ is equal to the product of all these primes. Setting $x = i$ we find $i^m = +i$ or $-i$ according as $m$ has an even or an odd number of prime factors, hence $i^m = \mu(m) \cdot i$ and $i^{2m} = (-1)^m = -1$. Thus $$ \sum_m \mu(m) \frac{i^m}{1-i^{2m}} = \frac i2 \cdot M, $$ where $M$ is the number of nonzero terms on the right.
On the other hand, since the number of primes is infinite, there must be infinitely primes $p \equiv 1 \bmod 4$, hence the left hand side is unbounded as $x \to i$. This is a contradiction.
Again I am certain that there are more clever ways of exploiting these identities to prove the infinitude of primes.
The product on the right consists of the sum of $\frac 1n$ where $n$ is any number divisible only by the primes $≤m$. These numbers may occur with some multiplicity, that doesn't matter. It is also easy to see that the product on the right is $$\frac 2{2-1}\times \frac 3{3-1}\times \cdots \times \frac m{m-1}=m$$
Thus, if every natural number from $1$ to $4^m$ factored completely using the primes $≤m$ we'd have that the left hand sum was $≤ m$ contrary to the stated premise.
Best Answer
Suppose there are only $m$ primes. The expression
$$\frac{p_i}{p_i-1} = \frac{1}{1-\frac{1}{p_i}} = 1 + \frac{1}{p_i} + \frac{1}{p_i^2} +\cdots,$$
by geometric series.
So if the right hand side is multiplied out, you get every possible unit fraction. So the right hand side must be equal to the harmonic series. But we've just shown that the harmonic series is strictly larger. So there must be more than $m$ primes.