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

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