Using ordered pairs and sequences to give a required condition for any counter-example to Legendre's Conjecture I found it very challenging to write this question.  I apologize for any ambiguity.
This is an argument that I am working on related to Legendre's Conjecture.  I appreciate any questions or any corrections.  :-)
Let:

*

*$x > 1$ be an integer

*$T(x)$ be the set of primes $p$ such that $0 < p < x$ and $p \nmid x(x+1)$

*$|T(x)|$ be the count of elements in $T(x)$

*$S_{T(x)}$ be a set of ordered pairs $(p_i,p_j)$ with the following properties:



*

*$p_i \ne p_j$

*$x^2 + x - p_i \equiv 0 \pmod {p_j}$

*$p_i,p_j \in T(x)$

Definitions:

*

*A sequence of primes $(p_1, p_2, p_3, p_4, \dots )$ is constructible from $S_{T(x)}$ if the following:



*

*the first element $c$ in the sequence can be any prime such that $c \in T(x)$

*the next element $n$ can be any prime where the ordered pair $(c,n) \in S_{T(x)}$

*the sequence consists of $2$ or more elements where for each $(p_i,p_{i+1})$, it follows that $(p_i,p_{i+1}) \in S_{T(x)}$ and for each $(\dots,p_j,p_{j+1}. \dots)$, it follows that $(p_j,p_{j+1},\dots) \in S_{T(x)}$


*

*The set $S_{T(x)}$ is said to have repeats if a sequence is constructible that includes a prime more than once.

Example 1: $x=16, T(x)=\{3,5,7,11,13\}, S_{T(x)} = \{ (5,3), (7,5), (11,3), (13,7) \}$
In this example, $S_{T(x)}$ does not have repeats
Example 2: $x=31, T(x)=\{3,5,7,11,13,17,19,23,29\}, S_{T(x)} = \{ (3,23), (5,3), (5,7), (7,5), (11,3), (13, 11), (17,3), (17,5), (17,13), (19,7), (23,3), (23,17), (23,19), (29,3) \}$
In this example, $S_{T(x)}$ has repeats since the sequence $(3,23,3)$ is constructible from $S_{T(x)}$.
Claim: If a counter example exists to Legendre's Conjecture for a given $x$, the set of ordered pairs $S_{T(x)}$ will necessarily have repeats.
(1)  Assume that for each $p \in T(x)$, there exists a $q \in T(X)$ such that $(p,q) \in S_{T(X)}$  [If this were not true, $x^2 + x - p$ would necessarily be a prime.]
(2)  Let $n = |T(x)|$ be the number of elements in $T(x)$.
(3)  We can arbitrarily order each $p \in T(x)$ up to $p_{n-1}$ in the following way:

*

*$p_1 = $ the least prime in $T(x)$.

*For each prime $p_i$, let $p_{i+1}$ be any prime where $(p_i,p_{i+1}) \in S_{T(x)}$ and $p_{i+1}$ has not yet been assigned an order.  [If it has already been assigned an order, then we have a repeat]

*Assume that we do not run out of primes (otherwise, we have reached a repeat and the argument is proven)

(4)  At $p_n$, there is no remaining unordered primes.  Since, by assumption, there exists $p_i \in T(x)$ such that $(p_n,p_i) \in S_{T(x)}$, it follows that there must be a repeat.
Does this argument hold?
Thanks.

Edit:
Great comment from John Omielan.  I have made changes to make the question less confusing.

*

*Added a definition for $x$

*Removed $2$ from examples since $2 | x^2 + x$ so it will never be an element of $T(x)$.

*Changed $p_1$ to $p_4$ in the definition of constructible to avoid any confusion.

*Add a specific $x$ value for the first example

*Removed some redundancies in the definition of $S_{T(x)}$

*Modified the title to make my question clearer.

*Updated Example 2 to make it complete

*Updated the third point in definition of constructible
 A: In your Definitions: section, you wrote:


*

*A sequence of primes $(p_1, p_2, p_3, p_4, \dots )$ is constructible from $S_{T(x)}$ if the following:



  
*
  
*the first element $c$ in the sequence can be any prime such that $c \in T(x)$
  
*the next element $n$ can be any prime where the ordered pair $(c,n) \in S_{T(x)}$
  
*the sequence consists of $1$ or more elements where the first element in the sequence meets the first condition and all other elements meet the second condition.
  


I found this quite confusing. Your third point states the sequence consists of $1$ or more elements. However, your second point requires there to be at least $2$ elements. Also, the last part of the third point says "all other elements meet the second condition". However, the second condition uses $c$, which is defined in the first condition for the first element. This seems to imply all of the other elements, say $p_i$, must have $(p_1,p_i) \in S_{T(x)}$. However, your second example stating $(3,23,3)$ is constructible shows this is not the case.
Instead, from the context used later in your Claim, it seems the definition can be just simply stated as:


*

*A sequence of $n \ge 2$ primes $(p_1, p_2, p_3, \dots, p_n)$ is constructible from $S_{T(x)}$ if $(p_i, p_{i+1}) \in S_{T(x)} \; \forall \; 1 \le i \le n - 1$.


Assuming this is what you mean, then regarding your Claim, any counter-example to Legendre's conjecture would not have any prime $p_L$ where $x^2 \lt p_L \lt x^2 + x$, although the claim itself is for $x^2 \lt p_L \lt x^2 + 2x + 1$. As such, your first part $(1)$ must hold since, as you state, otherwise if it did not, then $p_L = x^2 + x - p$ must be a prime. This is because if it were composite, it would have at least $2$ factors, with the smaller one being less than $x$. Any prime factor of this smaller factor can't divide $x^2 + x$ (if it did, then it must also divide $p$), so it must be in $T(x)$.
The rest of your Claim only depends on your part $(1)$ since, using it, you can create an unlimited length chain of primes in multiple ways (the rest of your claim gives one method, although note $p_1$ can actually be chosen to be any prime, not necessarily the smallest one) that, since there are only a finite number of primes in $T(x)$, must eventually repeat.
In conclusion, your stated Claim is true.
