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

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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

Aiming for a contradiction, suppose 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:

\(\ds T_n\) \(=\) \(\ds \dfrac {n \paren {n + 1} } 2\)
\(\ds T_{n + 1}\) \(=\) \(\ds \dfrac {\paren {n + 1} \paren {n + 2} } 2\)
\(\ds T_{n + 2}\) \(=\) \(\ds \dfrac {\paren {n + 2} \paren {n + 3} } 2\)
\(\ds T_{n + 3}\) \(=\) \(\ds \dfrac {\paren {n + 3} \paren {n + 4} } 2\)


Recall the definition of sphenic number:

A sphenic number is a (positive) integer which has exactly $3$ distinct prime factors all with multiplicity of $1$.


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 by hypothesis 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:

\(\ds T_x\) \(=\) \(\ds \frac {4 m \paren {4 m + 1} } 2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds 2 \times m \times \paren {4 m + 1}\)

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.

$\blacksquare$

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