This question is a follow-up to this earlier question.
Edit: (Updated (in response to this comment from mathlove - February 07, 2022 - 12:50 PM Manila time) It appears that we need the extra condition that $U$ is prime.
Edit: (Updated (in response to this comment from mathlove - February 07, 2022 - 01:08 PM Manila time) $U = 11$ is a counterexample, if we restrict $U$ to primes. I have asked mathlove to consider the case of a Mersenne prime $U = 2^t - 1$.
Let $U$ and $V$ be positive integers such that $U = V + 2$.
Here is my:
QUESTION
Can the positive integer $$T := T_U - T_V$$ be a triangular number, where $$T_X := \frac{X(X+1)}{2}$$ is the $X^{\text{th}}$ triangular number?
MY ATTEMPT #1
Plugging in $V = U - 2$ into the formula for $T$ and simplifying, we obtain $$T = \frac{U(U+1)}{2} - \frac{(U-2)(U-1)}{2} = 2U - 1.$$ Consequently, we obtain $$T \text{ is triangular } \iff 1 + 8T \text{ is a square, call it } S^2$$ $$\iff 1 + 8T = 1 + 8(2U - 1) = 16U - 7 = S^2.$$ The last equation implies that $S^2 \equiv 9 \mod {16}$. This is where I get stuck, as I do not currently know how to use the assumption that $U$ is a Mersenne prime to get a contradiction.
MY ATTEMPT #2
Since $U = 2^t - 1$ is a Mersenne prime, then $$U \equiv \begin{cases} { 3 \pmod {16}, t = 2 \\ 7 \pmod {16}, t = 3 \\ 15 \pmod {16}, t \geq 5. \\ } \end{cases} $$
Consequently, $$16U - 7 \equiv \begin{cases} { ({16}\cdot{3}) - 7 \equiv 9 \pmod {16}, t = 2 \\ ({16}\cdot{7}) - 7 \equiv 9 \pmod {16}, t = 3 \\ ({16}\cdot{15}) - 7 \equiv 9 \pmod {16}, t \geq 5, \\ } \end{cases} $$ whence we do not arrive at a contradiction.
$T_U - T_{U-2}$ can be a triangular number.
For $U=8$, one has $T_8-T_6=T_5$.
For $U=11$ which is prime, one has $T_{11}-T_9=T_6$.
For $U=127$ which is a Mersenne prime, one has $T_{127}-T_{125}=T_{22}$.