Fourth Power is Sum of 2 Triangular Numbers/Proof 2
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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
\(\ds T_{n^2 - n - 1} + T_{n^2 + n - 1}\) | \(=\) | \(\ds \frac {\paren {n^2 - n - 1} \paren {n^2 - n} } 2 + \frac {\paren {n^2 + n - 1} \paren {n^2 + n} } 2\) | 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 |
$\blacksquare$