Integer is Sum of Three Triangular Numbers

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Theorem

Let $n$ be a positive integer.

Then $n$ is the sum of $3$ triangular numbers.


Proof

From Integer as Sum of Three Squares, every positive integer not of the form $4^n \paren {8 m + 7}$ can be expressed as the sum of three squares.

Hence every positive integer $r$ such that $r \equiv 3 \pmod 8$ can likewise be expressed as the sum of three squares.


From Square Modulo 8, the squares modulo $8$ are $0, 1$ and $4$.

Thus for $r$ to be the sum of three squares, each of those squares needs to be congruent modulo $8$ to $1$.

Thus each square is odd, and $r$ can be expressed in the form $8 n + 3$ as the sum of $3$ odd squares.

So:

\(\, \displaystyle \forall n \in \Z_{\ge 0}: \, \) \(\displaystyle 8 n + 3\) \(=\) \(\displaystyle \paren {2 x + 1}^2 + \paren {2 y + 1}^2 + \paren {2 z + 1}^2\) for some $x, y, z \in \Z_{\ge 0}$
\(\displaystyle \) \(=\) \(\displaystyle 4 x^2 + 4 x + 4 y^2 + 4 y + 4 z^2 + 4 z + 3\)
\(\displaystyle \) \(=\) \(\displaystyle 4 \paren {x \paren {x + 1} + y \paren {y + 1} + z \paren {z + 1} } + 3\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n\) \(=\) \(\displaystyle \frac {x \paren {x + 1} } 2 + \frac {y \paren {y + 1} } 2 + \frac {z \paren {z + 1} } 2\) subtracting $3$ and dividing both sides by $8$


By Closed Form for Triangular Numbers, each of $\dfrac {x \paren {x + 1} } 2$, $\dfrac {y \paren {y + 1} } 2$ and $\dfrac {z \paren {z + 1} } 2$ are triangular numbers.

$\blacksquare$


Also known as

This theorem is often referred to as Gauss's Eureka Theorem, from Carl Friedrich Gauss's famous diary entry.


Historical Note

Carl Friedrich Gauss proved that every Integer is Sum of Three Triangular Numbers.

The $18$th entry in his diary, dated $10$th July $1796$, made when he was $19$ years old, reads:

$**\Epsilon\Upsilon\Rho\Eta\Kappa\Alpha \quad \text{num} = \Delta + \Delta + \Delta.$


Sources