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

Theorem
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.

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:

Recall the definition of sphenic number:

Hence:

That is:

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:

and similarly:

That is:
 * $\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$

Then we have:

There are three possibilities for $n$:

$\map \Omega n = 1$
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:

none of which are sphenic.

Thus:
 * $\map \Omega n \ne 1$

$\map \Omega n = 2$
Suppose $\map \Omega n = 2$.

We have:

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:
 * $\map \Omega n \ne 2$

$\map \Omega n = 3$
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 $n + 2 = 12$:

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:
 * $\map \Omega n \ne 3$

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.

Also see

 * Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic: $T_{28}$, $T_{29}$ and $T_{30}$ that is, $406$, $435$ and $465$, are the smallest $3$ consecutive triangular numbers which are all sphenic.


 * Sequences of 3 Consecutive Triangular Numbers which are Sphenic