Odd Square is Eight Triangles Plus One

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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.

$\blacksquare$


Illustration

8TrianglesPlus1.png


Also see


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.


Sources