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

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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 1

Let $\map \Omega n$ denote the number of prime factors of $n$ counted with multiplicity.


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.

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 $1$.


Hence:

\(\ds \forall m \in \closedint 0 3: \, \) \(\ds \map \Omega {T_{n + m} }\) \(=\) \(\ds 3\) that is, $T_{n + m} = p \times q \times r$ for some primes $p$, $q$, $r$
\(\ds \leadsto \ \ \) \(\ds \map \Omega {2 T_{n + m} }\) \(=\) \(\ds 4\) that is, $2 T_{n + m} = 2 \times p \times q \times r$
\(\ds \leadsto \ \ \) \(\ds \map \Omega {2 \times \dfrac {\paren {n + m} \paren {n + m + 1} } 2}\) \(=\) \(\ds 4\) Closed Form for Triangular Numbers
\(\ds \leadsto \ \ \) \(\ds \map \Omega {\paren {n + m} \paren {n + m + 1} }\) \(=\) \(\ds 4\) simplifying


That is:

\(\ds \) \(\) \(\ds \map \Omega {n \paren {n + 1} }\)
\(\ds \) \(=\) \(\ds \map \Omega {\paren {n + 1} \paren {n + 2} }\)
\(\ds \) \(=\) \(\ds \map \Omega {\paren {n + 2} \paren {n + 3} }\)
\(\ds \) \(=\) \(\ds \map \Omega {\paren {n + 3} \paren {n + 4} }\)
\(\ds \) \(=\) \(\ds 4\)


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:

\(\ds \map \Omega {n \paren {n + 1} }\) \(=\) \(\ds \map \Omega {\paren {n + 1} \paren {n + 2} }\) from above
\(\ds \leadsto \ \ \) \(\ds \map \Omega n\) \(=\) \(\ds \map \Omega {n + 2}\) Big Omega Function is Completely Additive


and similarly:

\(\ds \map \Omega {\paren {n + 2} \paren {n + 3} }\) \(=\) \(\ds \map \Omega {\paren {n + 3} \paren {n + 4} }\) from above
\(\ds \leadsto \ \ \) \(\ds \map \Omega {n + 2}\) \(=\) \(\ds \map \Omega {n + 4}\) Big Omega Function is Completely Additive

That is:

$\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$


Then we have:

\(\ds \map \Omega {\paren {n + 1} \paren {n + 2} }\) \(=\) \(\ds \map \Omega {\paren {n + 2} \paren {n + 3} }\) from above
\(\ds \leadsto \ \ \) \(\ds \map \Omega {n + 1}\) \(=\) \(\ds \map \Omega {n + 3}\) Big Omega Function is Completely Additive


There are three possibilities for $n$:

\(\text {(1)}: \quad\) \(\ds \map \Omega n\) \(=\) \(\ds 1\)
\(\text {(2)}: \quad\) \(\ds \map \Omega n\) \(=\) \(\ds 2\)
\(\text {(3)}: \quad\) \(\ds \map \Omega n\) \(=\) \(\ds 3\)


$\map \Omega n = 1$

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:

\(\ds T_3\) \(=\) \(\ds 6\) \(\ds = 2 \times 3\)
\(\ds T_4\) \(=\) \(\ds 10\) \(\ds = 2 \times 5\)
\(\ds T_5\) \(=\) \(\ds 15\) \(\ds = 3 \times 5\)
\(\ds T_6\) \(=\) \(\ds 21\) \(\ds = 3 \times 7\)

none of which are sphenic.

Thus:

$\map \Omega n \ne 1$

$\Box$


$\map \Omega n = 2$

Suppose $\map \Omega n = 2$.

We have:

\(\ds \map \Omega {n + 1}\) \(=\) \(\ds 2\) because $\map \Omega {n \paren {n + 1} } = 4$
\(\ds \map \Omega {n + 2}\) \(=\) \(\ds 2\) because $\map \Omega n = \map \Omega {n + 2}$
\(\ds \map \Omega {n + 3}\) \(=\) \(\ds 2\) because $\map \Omega { \paren {n + 2} \paren {n + 3} } = 4$
\(\ds \map \Omega {n + 4}\) \(=\) \(\ds 2\) because $\map \Omega {n + 2} = \map \Omega {n + 4}$

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:

$\map \Omega n \ne 2$

$\Box$


$\map \Omega n = 3$

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 $n + 2 = 12$:

\(\ds T_{10}\) \(=\) \(\ds 55\) \(\ds = 5 \times 11\) which is not sphenic
\(\ds T_{11}\) \(=\) \(\ds 66\) \(\ds = 2 \times 3 \times 11\) which is sphenic
\(\ds T_{12}\) \(=\) \(\ds 78\) \(\ds = 2 \times 3 \times 13\) which is sphenic
\(\ds T_{13}\) \(=\) \(\ds 91\) \(\ds = 7 \times 13\) which is not sphenic
\(\ds T_{14}\) \(=\) \(\ds 105\) \(\ds = 3 \times 5 \times 7\) which is sphenic

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:

$\map \Omega n \ne 3$

$\Box$


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.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Also see

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