# A short, unusual proof that there are infinitely many primes

Sam Northshield [1] came up with the following clever proof that there are infinitely many primes.

Suppose there are only finitely many primes and let P be their product. Then

The original publication gives the calculation above with no explanation. Here’s a little commentary to explain the calculation.

Since prime numbers are greater than 1, sin(π/p) is positive for every prime. And a finite product of positive terms is positive. (An infinite product of positive terms could converge to zero.)

Since p is a factor of P, the arguments of sine in the second product differ from those in the first product by an integer multiple of 2π, so the corresponding terms in the two products are the same.

There must be some p that divides 1 + 2P, and that value of p contributes the sine of an integer multiple of π to the product, i.e. a zero. Since one of the terms in the product is zero, the product is zero. And since zero is not greater than zero, we have a contradiction.

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[1] A One-Line Proof of the Inﬁnitude of Primes, The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), p. 466

## 9 thoughts on “A short, unusual proof that there are infinitely many primes”

1. Hankel

The classical proof, which says that $P+1$ must contain a prime factor other than the factors of $P$, is simpler. It doesn’t even need a contrapositive.

2. Andrei

But the proof simply boils down to:

1) For any n there is a prime p such that p divides n (used without proof)
2) Let P be the product of all primes. From 1, it follows that there is a p such that p divides 2P+1 (recognizable in the above, when we say that sin(pi(1+2P)/p)=0)
3) Since P is the product of all primes, p divides P (obvious and used in the above)
4) p does not divide 1 (used implicitly when we say that sin(pi/p) > 0)
5) 2,3,4 cannot be true simultaneously because of elementary properties of divisibility,

It does not use any special property of the sine.

3. It’s not obvious to me why must there be some p that divides 1 + 2P?

4. I need help with this one. Why is it guaranteed that there exist a p that divides (1 + 2P)?

5. …oh, because if not, then 1+2P would be a prime. I see it now.

6. So it’s just an embellished version of the P+1 can’t be prime argument. This feels like the number theory version of writing a c compiler in one line of regex. 😉

7. The abundance of clever proofs for this demonstrates something: the infinitude of the primes is not a very important fact, mathematically speaking. If it were, then it would be a requirement to even build up the theory behind these alternate proofs, and the proofs themselves would be circular.

8. Good explanation. Just an another approach of euclid’s.

9. José Antonio

Beautiful proof, as well as thouse by Fibonacci’s numbers and the proof of Erdös. Good job. Thank you.