# Square Numbers which are Sum of Sequence of Odd Cubes

## Theorem

The sequence of square numbers which can be expressed as the sum of a sequence of odd cubes from $1$ begins:

$1, 1225, 1 \, 413 \, 721, 1 \, 631 \, 432 \, 881, \dotsc$

The sequence of square roots of this sequence is:

$1, 35, 1189, 40 \, 391, \dotsc$

## Proof

We have that:

 $\ds 1225$ $=$ $\ds 35^2$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^5 \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3$ $\ds 1 \, 413 \, 721$ $=$ $\ds 1189^2$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^{29} \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3$

From Sum of Sequence of Odd Cubes we have:

$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$

Thus we need to find all $n$ such that $2 n^2 − 1$ is square.

This corresponds to the Pell's Equation $x^2 - 2 y^2 = -1$, which has the positive integral solutions:

$\begin {array} {r|r} x & y \\ \hline 1 & 1 \\ 7 & 5 \\ 41 & 29 \\ 239 & 169 \\ 1393 & 985 \\ \end {array}$

and so on.

By substituting $y = n$ and $x = \sqrt {2 n^2 - 1}$, we see that the products of $x$ and $y$ values give the required sequence of square roots.

$\blacksquare$