Odd Square is Eight Triangles Plus One

Theorem

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

Then $n$ is square iff $n = 8 m + 1$ where $m$ is triangular.

Proof

Follows directly from the identity:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 8 \frac {k \left({k + 1}\right)} 2 + 1\)	\(=\)	\(\displaystyle 4 k^2 + 4 k + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({2 k + 1}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

as follows:

• Let $m$ be triangular.

Then $\exists k \in \Z: m = \dfrac {k \left({k + 1}\right)} 2$ from Closed Form for Triangular Numbers.

From the above identity, $8 m + 1 = \left({2 k + 1}\right)^2$ which is an odd square.

• Let $n$ be an odd square.

Then $n = r^2$ where $r$ is odd.

Let $r = 2 k + 1$ so $n = \left({2 k + 1}\right)^2$.

From the above identity, $n = 8 \dfrac {k \left({k + 1}\right)} 2 + 1 = 8 m + 1$ where $m$ is triangular.

$\blacksquare$

Also see

• Odd Square Modulo 8