# My own proof

So here’s a proof of my own concoction. It’s decently similar to Euclid’s, but goofier. As I mentioned last time, I don’t like proof by contradiction. Ironically, this proof requires it.

Assume for the sake of contradiction that $p_1,p_2,\dots,p_n$ are the only primes. Let’s say they’re in that order. That is, $p_1. Let

$M=p_1\cdot p_2\cdot\dots\cdot p_{n-1}$.

Note, I’m not including the largest prime in the product. Similar to Euclid’s proof, if I look at $M+1$, it’s not divisible by any prime in the product of $M$, so $M+1$ must be divisible by $p_n$. But we can also look at $M-1$. This is also divisible by $p_n$. So in fact their difference, $(M+1)-(M-1)=2$ must be divisible by $p_n$. Thus, the largest prime is no larger than $2$, contradicting the primality of $17$.

Even though, on a technical level, there is no fundamental difference between these proofs, I seem to think that this one is somehow cleaner. I suppose this is because you don’t have to be able to factor to see it in action. Really, you don’t have to factor to do Euclid’s proof either, but it’s usually stated in such a way where you do need to factor.