# Is this proof of the infinitude of primes valid?

The current issue (May 2015) of the American Mathematical Monthly has a one-line proof that there are an infinite number of primes, and I don't see why it is correct.

Here is the proof:

If the set of primes is finite, then

$$0 < \prod\limits_{p} \sin\left(\frac{\pi}{p}\right) = \prod\limits_{p} \sin\left(\frac{\pi(1+2\prod_{p'}p')}{p}\right) =0 .$$

(That's the whole proof.)

I see why the first equality holds, since, if there are only a finite number of primes, $p \mid \prod_{p'}p'$ for all $p$.

But I do not see why the second equality ("$= 0$") holds. None of the terms in the product are zero, and, since there are only a finite number of them, the product is not zero.

So, do I not understand the proof, or is the proof incorrect?

Thank you.

• I really like the comments to Strants' answer. – marty cohen Jun 15 '15 at 18:00

We must have that $1+2\prod_{p'}p'$ is divisible by some prime $q$, so $1+2\prod_{p'}p' = kq$ for some integer $k$. But then, $$\sin\left(\frac{\pi(1+2\prod_{p'}p')}{q}\right) = \sin \pi k = 0$$ which gives the right-hand equality.
• @guest : This differs from "Euclid's proof made abstruse" in that Euclid's actual proof did not begin with an assumption that only finitely many primes exist: Euclid's proof was not by contradiction. Euclid showed that (rephrased into modern concepts) for every finite set $S$ of primes (which need not be the smallest $n$ primes for some $n$) the prime divisors of $1+\prod S$ are not in $S$. Thus $S$ can always be extended to a larger finite set of primes. Dirichlet and many later mathematicians erroneously wrote that Euclid's proof was by contradiction. That's a naked emperor. – Michael Hardy Jun 15 '15 at 11:56