Sum of Sequence of Cubes divides 3 times Sum of Sequence of Fifth Powers

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Theorem

$\displaystyle \sum_{i \mathop = 1}^n i^3 \divides 3 \sum_{i \mathop = 1}^n i^5$

where $\divides$ denotes divisibility.


Proof

\(\displaystyle 3 \sum_{i \mathop = 1}^n i^5\) \(=\) \(\displaystyle {T_n}^2 \paren {4 T_n - 1}\) Sum of Sequence of Fifth Powers
\(\displaystyle \) \(=\) \(\displaystyle k {T_n}^2\) where $k = 4 T_n - 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle {T_n}^2\) \(\divides\) \(\displaystyle 3 \sum_{i \mathop = 1}^n i^5\) Definition of Divisor of Integer
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{i \mathop = 1}^n i^3\) \(\divides\) \(\displaystyle 3 \sum_{i \mathop = 1}^n i^5\) Square of Triangular Number equals Sum of Sequence of Cubes

$\blacksquare$


Sources