Integer as Sum of Polygonal Numbers/Lemma 2
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Theorem
Let $n, m \in \R_{>0}$ such that $\dfrac n m \ge 1$.
Define $I$ to be the open real interval:
- $I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$
Then:
- For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.
Proof
We need to show that $\paren {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } - \paren {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}} > 4$ when $\dfrac n m \ge 116$.
Let $x = \dfrac n m - 1$.
Then:
\(\ds \paren {\frac 2 3 + \sqrt {8 \paren {\frac n m} - 8} } - \paren {\frac 1 2 + \sqrt {6 \paren {\frac n m} - 3} }\) | \(>\) | \(\ds 4\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \sqrt {8 x} - \sqrt {6 x + 3}\) | \(>\) | \(\ds \frac {23} 6\) |
To simplify calculations, we consider:
\(\ds \sqrt {8 x} - \sqrt {6 x + 3}\) | \(>\) | \(\ds 4\) | which is greater than $\dfrac {23} 6$ | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 8 x + 6 x + 3 - 2 \sqrt {48 x^2 + 24 x}\) | \(>\) | \(\ds 16\) | squaring Both Sides | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \sqrt {48 x^2 + 24 x}\) | \(<\) | \(\ds 7 x - \frac {13} 2\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 48 x^2 + 24 x\) | \(<\) | \(\ds 49 x^2 - 91 x + \frac {169} 4\) | squaring Both Sides | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x^2 - 115 x + \frac {169} 4\) | \(>\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \paren {x - 115} + \frac {169} 4\) | \(>\) | \(\ds 0\) |
which is true when $x \ge 115$.
Thus this condition is satisfied when $\dfrac n m \ge 116$.
$\blacksquare$
Sources
- Jan 1987: Melvyn B. Nathanson: A Short Proof of Cauchy's Polygonal Number Theorem (Proceedings of the American Mathematical Society Vol. 99, no. 1: pp. 22 – 24)