If n is Triangular then so is 49n + 6

Theorem

Let $n$ be a triangular number.

Then $49 n + 6$ is also triangular.

Proof

Let $n$ be triangular.

Then:

$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$

So:

\(\ds 49 n + 6\)

\(=\)

\(\ds 49 \frac {k \paren {k + 1} } 2 + 6\)

\(\ds \)

\(=\)

\(\ds \frac {49 k^2 + 49 k + 12} 2\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {7 k + 3} \paren {7 k + 4} } 2\)

which is triangular.

$\blacksquare$

Historical Note

David M. Burton, in his Elementary Number Theory, revised ed. of $1980$, reports that the result If $n$ is Triangular then so is $9 n + 1$ was published by Leonhard Paul Euler in $1775$.

He published this along with If $n$ is Triangular then so is $25 n + 3$ and If $n$ is Triangular then so is $49 n + 6$.

There is an obvious pattern.

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$