Triple of Triangular Numbers whose Pairwise Sums are Triangular

Theorem

The following triplet of triangular numbers has the property that the sum of each pair of them, and their total, are all triangular numbers:

$66, 105, 105$

Proof

Throughout we use Closed Form for Triangular Numbers, which gives that the $n$th triangular number can be expressed as:

$T_n = \dfrac {11 \times 12} 2$

We have:

\(\ds 66\)

\(=\)

\(\ds \frac {11 \times 12} 2\)

and so is triangular

\(\ds 105\)

\(=\)

\(\ds \frac {14 \times 15} 2\)

and so is triangular

Then:

\(\ds 66 + 105\)

\(=\)

\(\ds 171\)

\(\ds \)

\(=\)

\(\ds \frac {18 \times 19} 2\)

and so is triangular

\(\ds 105 + 105\)

\(=\)

\(\ds 210\)

\(\ds \)

\(=\)

\(\ds \frac {20 \times 21} 2\)

and so is triangular

\(\ds 66 + 105 + 105\)

\(=\)

\(\ds 210\)

\(\ds \)

\(=\)

\(\ds \frac {23 \times 24} 2\)

and so is triangular

$\blacksquare$

Historical Note

The result Triple of Triangular Numbers whose Pairwise Sums are Triangular was reported in Richard K. Guy's Unsolved Problems in Number Theory, 2nd ed. of $1994$ as having been provided by Charles Ashbacher.

Sources

• 1994: Richard K. Guy: Unsolved Problems in Number Theory (2nd ed.)
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $66$