# Odd Square is Eight Triangles Plus One

## Theorem

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

Then $n$ is square if and only if $n = 8 m + 1$ where $m$ is triangular.

## Proof

Follows directly from the identity:

 $\ds 8 \frac {k \paren {k + 1} } 2 + 1$ $=$ $\ds 4 k^2 + 4 k + 1$ $\ds$ $=$ $\ds \paren {2 k + 1}^2$

as follows:

Let $m$ be triangular.

Then from Closed Form for Triangular Numbers:

$\exists k \in \Z: m = \dfrac {k \paren {k + 1} } 2$

From the above identity:

$8 m + 1 = \paren {2 k + 1}^2$

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 that $n = \paren {2 k + 1}^2$.

From the above identity:

$n = 8 \dfrac {k \paren {k + 1} } 2 + 1 = 8 m + 1$

where $m$ is triangular.

$\blacksquare$

## Historical Note

This result, according to David Wells in Curious and Interesting Numbers in $1986$, was known to Diophantus of Alexandria.

However, David M. Burton in his Elementary Number Theory, revised ed. of $1980$, attributes the result to Plutarch, circa $100$ C.E.