# An unusual proof of the infinitude of primes

In thinking about the twin prime conjecture, I imagined a strategy for a possible proof. In order to evaluate the strategy, I asked myself if a simple form of it could be applied to the prime numbers. So the following is not meant to be a particularly efficient or elegant approach to proving the infinitude of primes, but a trial balloon to see whether the strategy might have weaknesses that I have not appreciated.

My questions are: Is the reasoning in this proof tight and valid? Also, to anyone's knowledge has this approach been reported before? If so, I would like to see links or references to discussion pertinent to this line of thought.

Assumption to be disproved: There are at least two prime numbers, and the number of primes is finite, i.e. $$\mathbb P =\{p_1,p_2,\dots, p_k\}$$ for some finite $$k>1$$. $$p_k$$ is the largest prime, and every number $$>p_k$$ is composite.

$$k>1 \Rightarrow p_k\#>p_k$$ where $$p_i\#$$ denotes the primorial function, the product of the first $$i$$ primes. For each prime number $$p_j \in \mathbb P$$, consider numbers of the form $$p_k\#+p_j$$. Such numbers are $$>p_k$$ and therefore composite; either they have more than one distinct prime factor, or they are a power of one prime.

Case 1: $$p_k\#+p_j$$ has more than one prime factor: We can see by inspection that one of those prime factors is $$p_j$$ itself. $$p_k\#+p_j=p_ap_jr$$ where $$r\ge 1$$ and may be any of a unit, prime or composite. $$p_ap_j\mid (p_k\#+p_j) \wedge p_ap_j\mid p_k\# \Rightarrow p_ap_j\mid (p_k\# +p_j - p_k\#) \Rightarrow p_ap_j\mid p_j$$ This is a contradiction of the primality of $$p_j$$, so Case 1 never applies and $$p_k\#+p_j$$ must be a power of a prime. As an aside, I note that $$p_ap_j\mid p_j \Rightarrow p_a \mid 1$$, which is the contradiction arrived at by a different route in Kummer's restatement of Euclid's proof. In the present instance, the equally impossible circumstance $$p_a \mid p_j$$ constitutes an independent contradiction.

Case 2: $$p_k\#+p_j$$ is a power of one prime: Here, $$p_j \mid p_k\#+p_j$$, so it must be the case that $$p_k\#+p_j=p_j^m$$. Since $$p_k\#$$ is a constant, it will be true for every $$p_j$$ that $$p_k\#+p_j = p_j^m \Rightarrow p_k\#=p_j^m-p_j$$ In particular, it must be true that $$p_k\#=2^a-2=3^b-3$$. This requires $$2^a=3^b-1$$. This special case of Catalan's conjecture was shown by Levi ben Gerson in the 14th century to have only $$a=3,\ b=2$$ as a solution (for $$a,b>1$$). This would imply that $$p_k\# = 6;\ k=2;\ \text{and}\ \mathbb P=\{2,3\}$$ This is plainly false, so either the assumption that there are at least two primes, or the assumption that the number of primes is finite, must be incorrect. But there are at least two primes, viz $$\{2,3,5,7,\dots\}$$. So the number of primes is infinite.

• My only doubt is whether the proof of Catalan's conjecture uses the fact that primes are infinite or not May 9, 2020 at 19:26
• I had that worry also (and I couldn't find a clear answer), which is why I went with Gerson's proof regarding so-called harmonic numbers, which is limited to consideration of numbers of the form $2^m3^n$. May 9, 2020 at 19:29
• @Ben W I don't see that Case 1 covers everything. If you think you can let $p_a=p_j$ in that case, it's not true that $p_j^2 \mid p_k\#$, so the further math doesn't follow. What do you see that you think I'm missing? May 9, 2020 at 20:13