Let $P_{1} = 2$, $P_{2} = 3$, $P_{3} = 5$, $P_{4} = 7$, ... (the list of all primes). Consider the function $F(x)$. It evaluates to $1$ if there exists some positive integer $N$ ( $N \ge 4$ ) such that
$$\left\{ \begin{array}{l} {P_{N - 2}} - {P_{N - 3}} \ne {P_{N - 1}} - {P_{N - 2}}\\ {P_N} - {P_{N - 1}} = {P_{N + 1}} - {P_N},\\ {P_{N + 2}} - {P_{N + 1}} = {P_{N + 3}} - {P_{N + 2}},\\ {P_{N + 4}} - {P_{N + 3}} = {P_{N + 5}} - {P_{N + 4}},\\ \cdots ,\\ {P_{N + 2(x - 2)}} - {P_{N + 2(x - 2) - 1}} = {P_{N + 2(x - 2) + 1}} - {P_{N + 2(x - 2)}},\\ {P_{N + 2(x - 1)}} - {P_{N + 2(x - 1) - 1}} = {P_{N + 2(x - 1) + 1}} - {P_{N + 2(x - 1)}},\\ {P_{N + 2x}} - {P_{N + 2x - 1}} = {P_{N + 2x + 1}} - {P_{N + 2x}},\\ {P_{N + 2x + 2}} - {P_{N + 2x + 1}} \ne {P_{N + 2x + 3}} - {P_{N + 2x + 2}} \end{array} \right.$$
(note that the total number of consecutive “ $=$ ”s between two “ $\ne$ ”s is always equal to $x+1$). Otherwise, $F(x) = 0$.
I tested first $100 000$ primes and found proofs that $F(x) = 1$ for all $x \in \{ 0,1,2\}$. For example, $F(2) = 1$ because there exists $N = 4209$ such that $P_{4209} = 40063$ and
$$\left\{ \begin{array}{l} {\rm{40037}} - {\rm{40031}} \ne {\rm{40039}} - {\rm{40037}}\\ {\rm{40063}} - {\rm{40039}} = {\rm{40087}} - {\rm{40063}},\\ {\rm{40093}} - {\rm{40087}} = {\rm{40099}} - {\rm{40093}},\\ {\rm{40111}} - {\rm{40099}} = {\rm{40123}} - {\rm{40111}},\\ {\rm{40127}} - {\rm{40123}} \ne {\rm{40129}} - {\rm{40127}} \end{array} \right.$$
where the total number of consecutive “ $=$ ”s between two “ $\ne$ ”s is equal to $3$.
I have not found any $N$ that would prove that $F(3) = 1$, let alone $F(4)$ etc.
Can we assume that $F(x) = 1$ for any positive $x$? If yes (or no), is it possible to prove this?
I think this follows from standard conjectures in Number Theory, but I doubt it follows from known results. Standard conjectures imply, for example, that there are infinitely many $n$ such that each of $n,n+210,n+420,n+840,n+1260,n+1470,n+1680,n+2100,n+2520$ is prime. These would be the nine primes to satisfy your four equations, in the case $x=3$. To make them consecutive primes would just require making the odd numbers between them composite, which is a finite number of congruences with solutions by the Chinese Remainder Theorem. Similarly, making the inequations you want would just be some more simple congruences.
But the smallest example might be very large.