Theorem: There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers

Let $n \in \N$ be a natural number.
Let $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ be the $n$th, $n + 1$th, $n + 2$th and $n + 3$th triangular numbers respectively.
Then it is not the case that all of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are sphenic numbers.

Proof:
Let $\map \Omega n$ denote the number of prime factors of $n$ counted with multiplicity.
{{AimForCont}} there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Thus from Closed Form for Triangular Numbers:
{{begin-eqn}}
{{eqn | l = T_n
      | r = \dfrac {n \paren {n + 1} } 2
}}
{{eqn | l = T_{n + 1}
      | r = \dfrac {\paren {n + 1} \paren {n + 2} } 2
}}
{{eqn | l = T_{n + 2}
      | r = \dfrac {\paren {n + 2} \paren {n + 3} } 2
}}
{{eqn | l = T_{n + 3}
      | r = \dfrac {\paren {n + 3} \paren {n + 4} } 2
}}
{{end-eqn}}
Recall the definition of sphenic number:
{{:Definition:Sphenic Number}}
Hence:
{{begin-eqn}}
{{eqn | q = \forall m \in \closedint 0 3
      | l = \map \Omega {T_{n + m} }
      | r = 3
      | c = that is, $T_{n + m} = p \times q \times r$ for some primes $p$, $q$, $r$
}}
{{eqn | ll= \leadsto
      | l = \map \Omega {2 T_{n + m} }
      | r = 4
      | c = that is, $2 T_{n + m} = 2 \times p \times q \times r$
}}
{{eqn | ll= \leadsto
      | l = \map \Omega {2 \times \dfrac {\paren {n + m} \paren {n + m + 1} } 2}
      | r = 4
      | c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadsto
      | l = \map \Omega {\paren {n + m} \paren {n + m + 1} }
      | r = 4
      | c = simplifying
}}
{{end-eqn}}
That is:
{{begin-eqn}}
{{eqn | o = 
      | r = \map \Omega {n \paren {n + 1} }
}}
{{eqn | r = \map \Omega {\paren {n + 1} \paren {n + 2} }
}}
{{eqn | r = \map \Omega {\paren {n + 2} \paren {n + 3} }
}}
{{eqn | r = \map \Omega {\paren {n + 3} \paren {n + 4} }
}}
{{eqn | r = 4
}}
{{end-eqn}}
Suppose one of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.
Then at least one of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ is divisible by $4$, and so is not sphenic.
Hence none of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.
We have that:
{{begin-eqn}}
{{eqn | l = \map \Omega {n \paren {n + 1} }
      | r = \map \Omega {\paren {n + 1} \paren {n + 2} }
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \map \Omega n
      | r = \map \Omega {n + 2}
      | c = Big Omega Function is Completely Additive
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = \map \Omega {\paren {n + 2} \paren {n + 3} }
      | r = \map \Omega {\paren {n + 3} \paren {n + 4} }
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \map \Omega {n + 2}
      | r = \map \Omega {n + 4}
      | c = Big Omega Function is Completely Additive
}}
{{end-eqn}}
That is:
:$\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$
Then we have:
{{begin-eqn}}
{{eqn | l = \map \Omega {\paren {n + 1} \paren {n + 2} }
      | r = \map \Omega {\paren {n + 2} \paren {n + 3} }
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \map \Omega {n + 1}
      | r = \map \Omega {n + 3}
      | c = Big Omega Function is Completely Additive
}}
{{end-eqn}}
There are three possibilities for $n$:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map \Omega n
      | r = 1
}}
{{eqn | n = 2
      | l = \map \Omega n
      | r = 2
}}
{{eqn | n = 3
      | l = \map \Omega n
      | r = 3
}}
{{end-eqn}}
=== $\map \Omega n = 1$ ===
Suppose $\map \Omega n = 1$.
Then $n$ is prime by definition.
Then as $\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$ that means they are all prime. 
But $n$, $n + 2$ and $n + 4$ cannot be all primes unless $n = 3$.
But we have:
{{begin-eqn}}
{{eqn | l = T_3
      | r = 6
      | rr= = 2 \times 3
      | c = 
}}
{{eqn | l = T_4
      | r = 10
      | rr= = 2 \times 5
      | c = 
}}
{{eqn | l = T_5
      | r = 15
      | rr= = 3 \times 5
      | c = 
}}
{{eqn | l = T_6
      | r = 21
      | rr= = 3 \times 7
      | c = 
}}
{{end-eqn}}
none of which are sphenic.
Thus:
:$\map \Omega n \ne 1$
{{qed|lemma}}
=== $\map \Omega n = 2$ ===
Suppose $\map \Omega n = 2$.
We have:
{{begin-eqn}}
{{eqn | l = \map \Omega {n + 1}
      | r = 2
      | c = because $\map \Omega {n \paren {n + 1} } = 4$
}}
{{eqn | l = \map \Omega {n + 2}
      | r = 2
      | c = because $\map \Omega n = \map \Omega {n + 2}$
}}
{{eqn | l = \map \Omega {n + 3}
      | r = 2
      | c = because $\map \Omega { \paren {n + 2} \paren {n + 3} } = 4$
}}
{{eqn | l = \map \Omega {n + 4}
      | r = 2
      | c = because $\map \Omega {n + 2} = \map \Omega {n + 4}$
}}
{{end-eqn}}
At least one of $n$, $n + 1$, $n + 2$, $n + 3$ and $n + 4$ is divisible by $4$.
But because $\map \Omega 4 = 2$, the only natural number divisible by $4$ is $4$ itself.
Thus one of $n$, $n + 1$, $n + 2$, $n + 3$ and $n + 4$ has to be equal to $4$.
We have already noted that none of $T_3$ to $T_6$ is sphenic.
Thus:
:$\map \Omega n \ne 2$
{{qed|lemma}}
=== $\map \Omega n = 3$ ===
Suppose $\map \Omega n = 3$.
From the above, we have:
:$\map \Omega {n + 1} = \map \Omega {n + 3} = 1$
and from $\map \Omega {n \paren {n + 1} } = 4$ we have:
:$\map \Omega {n + 1} = 1$
Thus $n + 1$ and $n + 3$ are twin primes.
Thus neither $n + 1$ nor $n + 3$ is divisible by $3$.
Thus $n + 2$ is divisible by $3$.
Suppose $n + 2$ is divisible by $12$.
Then as $\map \Omega {n + 2} = 3$, it follows that:
:$n + 2 = 12$
We investigate the case where $n + 2 = 12$:
{{begin-eqn}}
{{eqn | l = T_{10}
      | r = 55
      | rr= = 5 \times 11
      | c = which is not sphenic
}}
{{eqn | l = T_{11}
      | r = 66
      | rr= = 2 \times 3 \times 11
      | c = which is sphenic
}}
{{eqn | l = T_{12}
      | r = 78
      | rr= = 2 \times 3 \times 13
      | c = which is sphenic
}}
{{eqn | l = T_{13}
      | r = 91
      | rr= = 7 \times 13
      | c = which is not sphenic
}}
{{eqn | l = T_{14}
      | r = 105
      | rr= = 3 \times 5 \times 7
      | c = which is sphenic
}}
{{end-eqn}}
So $n + 2$ is not $12$ and so cannot be divisible by $12$.
But $n + 2$ is divisible by $2$, but not $4$.
Thus $n$ and $n + 4$ must be divisible by $4$.
Hence either $n$ or $n + 4$ will be divisible by $8$.
This contradicts our deduction that neither $n$ nor $n + 4$ can be divisible by $8$.
Thus:
:$\map \Omega n \ne 3$
{{qed|lemma}}
We have deduced that there exist no $n \in \N$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Hence the result by Proof by Contradiction.
{{qed}}