Fourth Power is Sum of 2 Triangular Numbers

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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.


Proof 1

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:

\(\ds T_{n^2 - 1} + T_{n^2}\) \(=\) \(\ds \frac {\paren {n^2 - 1} \paren {\paren {n^2 - 1} + 1} } 2 + \frac {\paren {n^2} \paren {n^2 + 1} } 2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \frac {n^2 \paren {n^2 - 1} } 2 + \frac {n^2 \paren {n^2 + 1} } 2\)
\(\ds \) \(=\) \(\ds \frac {n^2 \paren {2 n^2} } 2\)
\(\ds \) \(=\) \(\ds n^4\)

$\blacksquare$


Proof 2

\(\ds T_{n^2 - n - 1} + T_{n^2 + n - 1}\) \(=\) \(\ds \frac {\paren {n^2 - n - 1} \paren {n^2 - n} } 2 + \frac {\paren {n^2 + n - 1} \paren {n^2 + n} } 2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \frac {n \paren {n^2 - n - 1} \paren {n - 1} } 2 + \frac {n \paren {n^2 + n - 1} \paren {n + 1} } 2\) factoring
\(\ds \) \(=\) \(\ds \frac {n \paren {n^3 - n^2 - n - n^2 + n + 1} } 2 + \frac {n \paren {n^3 + n^2 - n + n^2 + n - 1} } 2\) multiplying out the numerators
\(\ds \) \(=\) \(\ds \frac {n \paren {\paren {n^3 - 2 n^2 + 1} + \paren {n^3 + 2 n^2 - 1} } } 2\) simplifying
\(\ds \) \(=\) \(\ds n^4\) simplifying

$\blacksquare$


Examples

$T_{41} + T_{55}$

$7^4 = T_{41} + T_{55}$


Historical Note

David Wells states in Curious and Interesting Numbers, 2nd ed. ($1997$) that this result was deduced by M.N. Khatri, but fails to give details.


Sources

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