Integer as Sum of Polygonal Numbers/Lemma 3
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Theorem
Let $n, m, r \in \R_{>0}$.
Suppose $\dfrac n m > 1$.
Let $b \in \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$.
Define:
- $a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$
Then $a, b$ satisfy:
- $b^2 < 4 a$
- $3 a < b^2 + 2 b + 4$
Proof
$b^2 < 4 a$
- $b^2 - 4 a = b^2 - 4 \paren {1 - \dfrac 2 m} b - 8 \paren {\dfrac {n - r} m}$
By the Quadratic Formula, $b^2 - 4 a < 0$ when $b$ is between:
\(\ds \) | \(\) | \(\ds \frac 1 2 \paren {4 \paren {1 - \frac 2 m} \pm \sqrt {16 \paren {1 - \frac 2 m}^2 + 32 \paren {\frac {n - r} m} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 - \frac 4 m \pm \sqrt {\paren {2 - \frac 4 m}^2 + 8 \paren {\frac n m} - 8 \paren {\frac r m} }\) |
Observing the term in the square root, we have:
- $2 - \dfrac 4 m - \sqrt {\paren {2 - \dfrac 4 m}^2 + 8 \paren {\dfrac n m} - 8 \paren {\dfrac r m} } < 0$
Since $b > 0$ this is satisfied.
Also we have:
\(\ds b\) | \(<\) | \(\ds \frac 2 3 + \sqrt {8 \paren {\frac n m} - 8}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 2 - \frac 4 m + \sqrt {8 \paren {\frac n m} - 8 \paren {\dfrac r m} }\) | $m \ge 3$, $r < m$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 - \frac 4 m + \sqrt {\paren {2 - \frac 4 m}^2 + 8 \paren {\frac n m} - 8 \paren {\dfrac r m} }\) | Square is nonnegative |
showing that first inequality is satisfied.
$\Box$
$3 a < b^2 + 2 b + 4$
- $b^2 + 2 b + 4 - 3 a = b^2 - \paren {1 - \dfrac 6 m} b - 6 \paren {\dfrac {n - r} m} + 4$
By the Quadratic Formula, $b^2 + 2 b + 4 - 3 a > 0$ when $b$ is greater than:
\(\ds \) | \(\) | \(\ds \frac 1 2 \paren {1 - \frac 6 m + \sqrt {\paren {1 - \frac 6 m}^2 + 4 \paren {6 \paren {\frac {n - r} m} - 4} } }\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \frac 1 2 \paren {1 + \sqrt {24 \paren {\frac n m} - 16} }\) | $m > 0$, $r \ge 0$, $-1 \le 1 - \dfrac 6 m < 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \sqrt {6 \paren {\frac n m} - \frac {15} 4}\) |
showing that second inequality is satisfied by any $b > \dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}$.
$\blacksquare$
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)