Fourth Power is Sum of 2 Triangular Numbers/Proof 1

Theorem

Let $n \in \Z$ be an integer.

Then:

$\exists a, b \in \Z_{\ge 0}: n^4 = T_a + T_b$

where $T_a$ and $T_b$ are triangular numbers.

That is, the $4$th power of an integer equals the sum of two triangular numbers.

Note first that:

$\forall n \in \Z: \paren {-n}^4 = n^4$

Hence it is sufficient to consider the case where $n \ge 0$.

For $n = 0$:

$0^4 = 0 = 0 + 0 = T_0 + T_0$

Let $n > 0$.

Then:

$n^2 - 1 \ge 0$

and:

$n^2 \ge 0$

Hence:

$\ds T_{n^2 - 1} + T_{n^2}$

$=$

$\ds \frac {\paren {n^2 - 1} \paren {\paren {n^2 - 1} + 1} } 2 + \frac {\paren {n^2} \paren {n^2 + 1} } 2$

$\ds $

$=$

$\ds \frac {n^2 \paren {n^2 - 1} } 2 + \frac {n^2 \paren {n^2 + 1} } 2$

$\ds $

$=$

$\ds \frac {n^2 \paren {2 n^2} } 2$

$\ds $

$=$

$\ds n^4$

$\blacksquare$