Gaps between primes

Question

I recently watched a video about the recent breakthrough involving the gaps between primes. I have an idea that I'm sure is wrong, but I don't know why.

1. If you take the product of all prime numbers up to a certain number and call it x, won't x-1 and x+1 always be primes?
2. And since they always differ by 2, doesn't that make there an infinite number of primes that differ by 2?

Once again, I know that I'm wrong, but I would like to know why.

Answer 1

Say $p_k$ is the largest prime $\le$ some number $n$. Then take $\displaystyle x=p_1p_2\cdots p_k$ Now, of course $x-1$ and $x+1$ are not divisible by any of $p_i,\ 1\le i\le k$. But there may be $p_k<p_i< x-1$ and $p_k<p_j<x+1$ such that $p_i|x-1$ and $p_j|x+1$. Just look at the example given by @DanielFischer to appreciate this fact.

Answer 2

What you propose is actually part of a very simple and elegant proof, proposed by Euclid (about 300 BC).

To prove that there are infinitely many primes:

• Suppose that $p_1 < p_2 < \dots < p_n$ are the only primes.
• Let $P = p_1p_2 \dots p_n + 1$. Now $P$ can be prime, but that is not the point.
• If $P$ is prime indeed, well, we have found another prime, because $P$ is not divisible by any of the primes $p_1 \dots p_n$. (The remainder in such case will always be one.)
• If $P$ is not prime, that means it has got a divisor $q$ that is neither $1$ nor $P$ itself. If $q$ was one of the numbers $p_1 \dots p_n$, that would mean it divides the product of all primes $p_1p_2 \dots p_n$. But we supposed that $q$ divides $P = p_1p_2 \dots p_n + 1$. Then $q$ has to divide the difference of both numbers (see below), which is $1$. And no prime can do that. Our assumption led us to contradiction.
• In either case, we have found another prime. This can be repeated infinitely many times, yielding infinitely many prime numbers.

Most people when they see this for the first time remember that $P$ is prime somehow and then get confused about it.

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Say $m,n$ have a common divisor $q$. Then we can write $m = uq$ and $n = tq$. Their difference $m - n$ can then be written as $d = uq - tq = (u - t)q$. As we can see, $d$ is divisible by $q$ as well.

Furthermore, Euclid's proof shows that such prime number $P$ exists, it does not "construct" it. We have shown that an abstract mathematical entity exists without actually finding it.

Finding huge primes quickly is extremely important for cryptography. A basic algorithm for finding primes is called the Sieve of Eratosthenes.

Answer 3

Others have already pointed out that the product of the fist $n$ primes plus one will not in general be a prime number.

Just to point out what Yitang Zhang actually proved (I assume that this is what you give referemce to).

Let $p_n$ be the $n$th prime number. Then it is true that $$ \liminf_{n\to \infty} p_{n+1} - p_n < 70\cdot 10^6. $$ That is, you will always be able to find conseqtive prime numbers that are at most 70 million apart.