$p_{i + k} - p_i \neq \text{const}$ for any $k \geq 1$ where $p_i = i$th prime number.

Question

Let $p_i$ be the $i$th prime number. This should be simple to prove:

$$ \forall k \geq 1, c \in \Bbb{Z}, \\ p_{i + k} - p_i \neq c, \\ $$ for some $i \geq 2$. But for example:

$$ 11 - 5 = 6 \\ 13 - 7 = 6 \\ 17 - 11 = 6 \\ 19 - 13 = 6 \\ 23 - 17 = 6 \\ $$

So long strips do exist, I'm saying the whole diagonal strip (a diagonal line in the table for $p_i - p_j$) cannot be constant.

I specified $i \geq 2$ because I'm only interested in odd primes, but lowering to $i \geq 1$ should also work.

It should be easy to prove, but I'm stuck on it.

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If $p_{i + k} - p_i = c$ for all $i \geq 2$ then in particular $3 + cz$ is prime for all $z \geq 0$, which would include $z \in 3 \Bbb{Z}$? Does that work as a proof?

My next question is how can we generalize the statement? I got a little too specific with the indices.

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I think I've got it now.

Let $b,c,k \in \Bbb{N}$ be positive numbers. Then:

$$ p_b + c = p_{b + k} \\ p_{b + k} + c = p_{b + 2k} \\ p_{b + 2k} + c = p_{b + 3k} \\ \vdots $$

would imply $$ p_b + c = p_{b + k} \\ p_b + 2c = p_{b + 2k} \\ p_b + 3c = p_{b + 3k} \\ \vdots $$ or in general $p_b + cz = p_{b + zk}$ for all $z \geq 0$. Thus choose any $z \in p_{b}\Bbb{N}$. For more tightness we'd only have to choose the minimal composite in $w \in p_b + c \Bbb{N}$, and then take $z = \dfrac{w - p_b}{c}$

Thus to generalize, $b + zk$ was in the place of $i + k$ so:

$$ b + zk = i + k = b + \dfrac{w - p_b}{c} k \implies \\ \exists i, b \leq i \leq b + k(\dfrac{w - p_b}{c} - 1) : \\ p_{i + k} - p_i \neq c \\ $$

Answer 1

If I understand your question, the fact that there there are, for any $n$, more than $n$ consecutive composite integers should prove what you want by taking $n > c$.

Answer 2

It's a standard argument that any non-constant polynomial $g(n) \in \mathbb Z[n]$ takes (arbitrarily large) composite values. Your problem is just the case of a polynomial of degree $1$.

Pick any value of $g(n_0)$ that you like (pick $n_0$ large enough so that $|g(n_0)|>1$). If it's composite, then great. If it's a prime $p$, then note that $g(n_0 + kp) \equiv g(n_0) \equiv 0 \pmod p$ for any $k\ge 1$. Now it's possible that $g(n_0 + kp)$ is exactly equal to $\pm p$ and hence still prime, but this can only happen for finitely many values of $k$. Thus $g(n_0 + kp)$ will eventually be always composite for large enough $k$.