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:

$$ $$

as follows:


 * Let $$m$$ be triangular.

Then $$\exists k \in \Z: m = \frac {k \left({k + 1}\right)} 2$$ 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, $$n = 8 \frac {k \left({k + 1}\right)} 2 + 1 = 8 m + 1$$ where $$m$$ is triangular.

Also see

 * Odd Square Modulo 8