There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers/Proof 2

Proof
there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.

Observe from Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic that there are no such $n$ for $n < 12$.

Thus from Closed Form for Triangular Numbers:

Recall the definition of sphenic number:

Let $S = \set {n, n + 1, n + 2, n + 3, n + 4}$.

Then there exists $x \in S$ such that $4 \divides x$.

If $x = n$ or $x = n + 4$, then one of $n$ or $n + 4$ is divisible by $8$.

Then one of $T_n$ or $T_{n + 3}$ is divisible by $4$, so it cannot be sphenic.

Therefore $x \ne n$ and $x \ne n + 4$.

Hence $x - 1 \in S$ and $x + 1 \in S$.

Note that both $T_{x - 1}$ and $T_x$ are sphenic.

Write $x = 4 m$ for some integer $m$.

Since $x > n \ge 12$, we have $m > 3$.

Note that:

so as $T_x$ is sphenic, both $m$ and $4 m + 1$ must be prime.

Similarly, since $T_{x - 1}$ is sphenic, $4 m - 1$ must be prime.

One of $4 m - 1, 4 m, 4 m + 1$ is divisible by $3$.

As $4 m + 1 > 4 m - 1 > 3$, $4 m$ must be divisible by $3$.

By Euclid's Lemma, since $3 \perp 4$:
 * $3 \divides m$

which contradicts the fact that $m$ is prime.

Hence the result by Proof by Contradiction.